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5 


6 


^^0^s^^m^^~ 


^^^.^"^i^.M''^'^''^'' 


2nd  COPY, 
1898. 


HEMATICAL   WORKS. 
ESSOR   EDWARD  A.    BOWSER. 


ACADEMIC  ALGEBRA.    With  Numerous  Examples. 

COLLEGE  ALGEBRA.    With  Numerous  Examples. 

PLANE  AND  SOLID  GEOMETRY.  With  Numerous 
Exercises. 

ELEMENTS  OF  PLANE  AND  SPHERICAL  TRIG- 
ONOMETRY.   With  Numerous  Examples. 

A  TREATISE  ON  PL/INE  AND  SPHERICAL  TRIG- 
ONOMETR^  ,  AND  ITS  Applications  to  Astronomy 
AND  Geodbsv.    With  Numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC 
GEOMETRY,  Embracing  Plans  Geometry  and 
AN  Introduction  to  Geometry  of  Three  Dimen- 
sions.   With  Numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  THE  DIFFER- 
ENTIAL AND  INTEGRAL  CALCULUS.  With 
Numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  ANALYTIC 
MECHANICS.    With  Numerous  Examples. 

AN  ELEMENTARY  TREATISE  ON  HYDROME- 
CHmNICS.    With  Numerous  Examples. 

LOGARITHMIC  AND   TRIGONOMETRIC  TABLES. 

A  TREATISE  ON  ROOFS  AND  BRIDGES.  With 
Numerous  Exercises. 


J; 


''' 


II  iiri(iir»iiiiii«ia»»!P''l'K"<!*''"r*B*'*'liW 


A  TREATISE 


RQOFS  AND  BRIDGES 


I    - 


WITH  NUMEROUS  EXERCISES 


BT 


/ 


EDWARD  A.   BOWSER 

PROFESSOR  OP  MATHKMA'ifrS  AND    ENOINEKRINO   IN    RUTQRRB   COLLBOR 


NEW  YORK 

D.   VAN   NOSTRAND   COMPANY 

23  Murray  and  27  ^;rAKRBN  Strkkts 

1898 


if 


.'i>mtuij  nil II    I   .niii^ifi.iii  1,1    HI  I, ui nil      u%if\iiri0iiif)im^iiiriifgimrmiifmi'>^ 


20013 


OoPTBisnT,  1898,  , 

Bt  e,  a.  bowser. 


OF 


DEC7-1CG8  J 


* 


I 


Xartnaati  )^\m 

1.  8.  Ctuhinx  Ii  Co       llprwick  .1  Smith 

Nurwoud  Mua8.  II.8.A 


f 


PREFACE. 


\^ 


r 


I 


Thk  present  treatise  on  Roofs  and  Bridges  is  designed  as 
a  textrbook  for  the  use  of  schools.  The  object  of  tliis  work 
is  ^o  develop  the  principles  and  explain  the  methods  em- 
ployed in  finding  the  forces  in  Roofs  and  Bridges,  and  to 
train  the  student  to  compute  the  stresses,  due  to  the  dead, 
live,  snow,  and  wind  loads,  in  the  different  members  of  any 
of  the  simple  roof  and  bridge  trusses  that  are  in  common 
use. 

The  aim  has  b<!en  to  explain  the  principles  clearly  and 
concisely,  to  develop  the  different  methods  simply  and 
neatly,  and  to  present  the  subject  in  accordance  with  the 
methods  used  in  the  modern  practice  of  roof  and  bridge 
construction. 

In  introducing  each  new  triiss  it  is  at  first  carefully  de- 
scribed, and  the  method  of  loading  it  explained.  A  prob- 
lem is  then  given  for  this  truss,  and  solved  to  determine  the 
stresses  in  all  the  members.  This  problem  is  followed  by 
several  other  similar  ones,  which  are  to  be  solved  by  the 
student.  Nearly  all  of  these  problems  were  prepared  espe- 
cially for  this  work,  and  solved  to  obtain  the  answers.  The 
instructor  can  at  any  time  easily  make  up  problems  for  his 
pupils  without  the  answers. 

The  hook  consists  of  four  chapters.  Chapter  I.  is  entirely 
given  to  Roof  Trusses.  Chapter  II.  treats  only  of  Bridge 
Trusses  with  Uniform  Loads. 

ili 


I  > 


iVBi 


Iv 


PREFACE. 


Chapter  III  is  devoted  to  IJridge  Trusses  with  Unequal 
Distribution  of  the  Loads.  This  is  divided  into  three  parts, 
as  follows : 

(1)  The  use  of  a  unifomihj  distributed  excess  load  covering 
one  or  mcjre  panels,  followed  by  a  uniform  train  load  cover- 
ing the  whole  span. 

(2)  The  use  of  one  or  two  concentrated  excess  loads,  with 
a  uniform  train  load  covering  the  span. 

(3)  The  use  of  the  actual  specijied  locomotive  ivheel  loads, 
followed  by  a  uniform  train  load. 

Chapter  IV  treats  of  Miscellaneous  Trusses,  including  the 
Crescent  Roof  Truss,  the  Pegram  and  Parabolic  Bowstring 
Uridge  Trusses,  and  Skew  Bridges. 

The  stresses  in  this  work  are  nearly  all  given  in  tons,  the 
word  "  ton  "  meaning  a  ton  of  2000  pounds.  Any  ocher 
unit  of  load  and  of  stress  might  be  used  as  well. 

My  best  thanks  are  due  to  my  friend  and  former  pupil, 
Mr.  George  H.  Blakeley,  C.E.,  of  the  class  of  '84,  now  * 
Chief  Engineer  of  the  Passaic  Rolling  Mill  Company,  for 
reading  the  manuscript  and  for  valuable  suggestions. 


E.  A.  B. 


RUTOBRS  COLLBOE, 

Nbw  Brunswick,  N.J.,  October,  1898. 


k 


*•»«> 


ii.M*«>»iSSa»^  -  ».i3£4&S&^;ri^^ 


ABT. 
1. 

2. 

3. 

4. 

5. 

(J. 

7. 

8. 

9. 
10. 
11. 
12. 


TABLE  OF  CONTENTS. 


CHAPTER  I. 
Roof  Tkuhses. 


Definitions        .... 
'I'lie  Dead  Load 
Tlio  Live  I-oad .... 
Tlie  Apex  F.oud.'i  and  Reactions 
Relations  between  External  Forces 
Aletliods  of  Calculation     . 
Lever  Arms  —  Indeterminate  Cases 
Snow  Load  Stresses  .        .        . 
Wind  Loads      .... 
Wind  A}iex  Loads  and  Reactions 
\Vind  Stresses  .... 
Complete  Calculation  cf  a  Roof  Truss 


and  Internal  Stresses 


1 
3 
4 
4 

0 
11 
14 
19 
20 
22 
25 
29 


aW 


CHAPTER  n. 
Bridge  Trusses  with  Uniform  Loads. 


13.  Definitions        .        .       .        . 

14.  Different  Forms  of  Trusses 

15.  The  Dead  Load 

10.  The  Live  Load  .... 

17.  Shear  —  Shearing  Stress   . 

18.  Web  Stresses  due  to  Dead  Loads 

19.  vbord  Stresses  due  to  Dead  Loads 


38 
89 
42 
44 
40 
47 
50 


A"iWl".' 


Vi 

AST. 

'JO. 

'Jl. 
22. 
23. 
24. 

2."). 
20. 
27. 
28. 

•jn. 
m. 

.12. 
:t:t. 
:tl. 

:<(!. 
;(7. 

38. 


CONTENTS. 

Position  of    IJiiifonii   Live    Loail    causiiif; 

(.'liord  Sti'0.ssc'N     .... 
Mftxiinum  Stresses  in  tiie  Clionls    . 
I'ositiuii  of  Uniform  I<ivo  Load  causing  Mt^xini 
Tiie  VVarit'ii  Truss    .... 
Main.s  and  (.'ounters. 
Tiio  Howe  Truss       .... 
Tlie  Pratt  Truss        .... 
TIni  WarriMi  Truss  witli  Vertical  Suspenders 
Tiie  Doulile  Warren  Truss 
Tlio  Wliipple  Truss  .... 


I 


Maximum 


The  Lattice  Truss 

The  Post  'J .  U8S 

The  Hollnum  Truss  . 

The  V'mk  Truss 

Tlie  Parabolic  Howstring  Truss 

The  Circular  Howstring  Truss 

Snow  Load  Stresses  . 

Stresses  due  to  Wind  Pressure. 

The  Factor  of  Safety 


nil  Shears 


54 

65 

67 

60 

70 

74 

81 

84 

87 

00 

04 

08 

100 

101 

103 

108 

lOi) 

110 

118 


^f 


30. 

40. 

41. 

42. 

43. 
44. 
45. 


CHAPTER  in. 

Bridge  TuussEa  with  Unhqual  Distridutiox 
OF  TiiK  Loads. 

Preliminary  Statement ii>o 

W'.ieu  the  Uniform  Train  Load  is  preceded  by  One  or 

More  Heavy  Excess  Panel  Loads  ....  120 
When  One  Concentrated  Excess  Load  accompanies  a 

Uniform  Train  I^cad 127 

When  Two  Ecjual  Concentrated  Excess  Loads  accompany 

a  Uniform  Train  Load l!J3 

The  Baltimore  Truss 142 

True  Maximum  Shears  for  Uniform  Live  Load      .        .  154 

Locotnotive  Wlieel  Loads 157 


~4l'M-f, 


'■■^■K~ 


M)>fHWMHH>>~.~,. 


51 
5fi 
R7 
60 
70 
74 
81 
84 
87 
00 
»1 
08 
100 
101 

lo:) 

108 
101) 

no 

118 


i..r 


^■f 


40.     PoHition  of  Wlusel  Loads  for  Mivxi'iium  Slicnr 

47.  I'.)sitioii   of   Wheel    I.oitdH   fur   Maxiiiiuiii    Moini'iit   at 

•loiiit  ill  Loudc'l  Chord  

48.  Posilioii  of   Wheel    I.oadH   for    Miixiiiiiiin    Moineiit   al 

Joint  ill  I'liloailrd  Chord 

40.    Tabulation  of  Mouiuuts  of  Wlieel  Loads. 

CIIAPTEll   IV. 

MiSCKLLAJJEOt'S   TllUSSKH. 

Ron/  Truttts. 

50.    The  King  and  Queen  Truss  —  The  Fink  Truss 

•A.    The  Crescent  Truss 


1(10 
108 


Bridge  Trusses. 

52.  'he  Pegram  Tniss  —  The  Parabolic  Bowstrinjr  Truss 

53.  tjKew  Bridges 


177 
182 


187 
102 


i;>o 


120 


127 


'I? 


1!5:j 

142 

154 
157 


1\  .^ 


-^iS^Wi'^- 


ROOFS   AND   BRIDGES. 


o»{o 


CHAFTER   I. 


ROOF  TRUSSES, 


Art.  1.    Definitions.  —  A  Tramed  Scructurs  is  a 

collection  of  pieces,  either  of  wood  or  ii-etal,  or  both 
combined,  so  joined  together  a.s  to  causn  the  structure  to 
act  as  one  rigid  body.  The  points  at  which  the  pieces  are 
joined  together  are  usually  c&V  ""d  joints. 

A  Truss  1?  a  structure  des  .^ned  to  transfer  loads  on  it 
to  the  supports  at  each  end,  while  each  member  of  the  truss 
is  subject  only  to  longitudinal  stress,  either  tension  or  com- 
pression. The  simplest  of  all  trusses  is  a  triangle;  and  all 
tmsses,  however  complicated,  containing  no  superfluous 
members,  must  be  composed  of  an  assemblage  of  triangles, 
since  a  triangle  is  the  only  polygon  whose  form  cannot  be 
changed  without  changing  the  lengths  of  its  sides. 

A  Strut  is  a  member  which  takes  compression.  Struts 
are  sometimes  called  j9os<t,  or  co?«mns. 

A  Tie  is  a  member  which  tivkes  tension. 

A  Brace  is  a  term  used  to  denote  both  struts  and  ties. 

A  Counter  Brace  is  a  member  which  Is  designed  to  take 
both  compressi'*u  and  tension.  A  Counter  is  a  member 
designed  to  take  eiclier  compression  or  tension  ;  that  is,  for 
one  position  of  the  load  the  member  may  be  compressed, 
while  for  another  position  it  may  be  elongated. 

1 


1 1 


2  HOOFS  AND  niilDGES. 

The  Uppev  and  Lower  Chords  are  the  upper  and 
l(nvcr  members  of  a  truss,  exti'iidiiig  from  one  support  to 
the  other.  Each  half  of  the  upjier  cl.iord  of  a  roof  truss 
is  sometimes  called  the  "main  rafter,"  while  the  lower 
chord  is  often  called  the  "tie  rod."  The  ujiper  chord  is 
always  in  compression  and  the  lower  chord  is  always  in 
tension.  The  spaces  between  tlie  join^^^s  of  the  chords  are 
called /)aHe?s. 

The  Web  Members  are  those  which  connect  the  joints 
of  the  two  chords.  They  are  generally  alternately  struts 
and  ties. 

A  Roof,  in  common  language,  is  the  covering  over  a 
stiucture,  the  chief  ol)ject  of  which  is  to  in-otect  the  build- 
ing from  the  effects  of  rain  and  snow. 

A  Roof  Truss  is  a  structure  which  suppoits  a  roof 
Roof  Trusses  are  of  almost  innumerable  forms,  and  they 
differ  greatly  in  the  details  of  their  construction. 

The  External  Forces  include  all  the  exterior  or  applied 
forces,  such  as  the  weight  of  the  structure,  the  weight  of 
snow,  the  force  of  the  wind,  the  reactions,  etc.,  which  act 
upon  and  tend  to  distort  the  structure.  The  external  forces 
oji  tlie  whole  structure  must  balance  each  other,  or  else  the 
whole  structure  Avill  begin  to  move. 

Strain  is  a  change  in  the  length  or  in  the  form  of  a  body, 
Avhich  has  been  produced  by  the  application  of  one  or  more 
external  forces;  and  it  is  to  be  measured,  not  in  tons,  but  ia 
units  of  length,  as  inches  or  feet. 

Stress  is  the  name  given  to  that  internal  force  which  is 
exerted  by  the  material  in  resisting  strain;  and  it  is  meas- 
ured in  pounds  or  tons  the  same  as  the  external  forces.  It 
follows  that,  when  every  part  of  a  strained  meu'ber  of  a 
structure  is  in  e(piilibrium,  the  internal  stress  exerted  at 
any  imaginary  section  through  the  member  is  equal  and 
opposite  to  the  straining  force. 


-'  : 


ROOF  TRUSSES. 


Various  Kinds  of  Stresses.  —  Tlio  external  forces  may 
prodiux',  accordiiij,'  to  eircumstaiices,  diHereut  internal  forces 
or  stresses  in  the  various  pieces  of  the  structure.  These 
forces  or  stresses  and  their  accompanying  strains  may  be 
classiiied  as  follows : 

1.  A  direct  pull  or  a  teiisilo  stress,  producing  exiension 
or  elongation. 

2.  A  direct  thrust  or  a  compressive  stress,  i)roducing 
compression. 

3.  A  sliearing  force  or  a  shearing  stress,  producing  a 
cutting  asunder. 

A  shearing  force  or  shearing  stress  is  caused  by  two  forces 
a<;ting  parallel  to  each  other,  and  at  right  angles  to  the  axis 
of  the  piece,  but  in  opposite  directions,  and  in  two  imme- 
diately consecutive  sections.  The  tendency  is  to  cause  two 
adjacent  sections  to  slide  o'.i  each  other;  and  the  term 
shearing  force,  or  shearing  strens,  is  iised^  l)ecause  the 
force  is  similar  to  that  in  the  blades  of  a  pair  of  shears  in 
the  act  of  cutting.  I   ;   ,;  i-      ,'    .      "   "-       <■  '> 

Art.  2.  The  Dead  Load.  —  The  dead  or  iixcd  load  sup- 
ported by  a  roof  truss  consists  of  the  weight  of  the  truss 
itself  and  the  weight  of  the  roof.  The  weight  of  the  truss 
can  .only  be  approximated.  The  following  formula*  gives 
approximately  correct  results  for  short  spans :  Let  I  =  span 
in  feet;  b  =  distance  between  trusses  in  feet;  and  W=  ap- 
proximate weight  of  one  truss  in  pounds.     Then 

The  roof  includes  the  roof-coverinf,  the  sheeting,  the 
rafters,  and  the  purlins.  Its  total  weight  will  vary  from 
6  to  30  lbs.  per  square  foot  of  roof  surface.     The  purlins  are 

•  Modern  Framed  Structures,  by  Johnson,  Bryan,  and  Turneaure. 


-^v«<i«^Mt«wv- 


*  <* 


•^^  --  #*1*^«»  .^«*w 


•  , 


!    . 


4  HOOFS  AND  BRIDGES. 

beams  ruiiuing  longit)idiiially  between  the  trusses,  and  fast 
ened  to  them  at  the  upper  joints,  and  often  midway  between 
them  as  well. 


I 


Art.  3.  The  Live  Load.  —  The  live  or  variable  load 
consists  of  the  snoiv  load  and  the  wind  loiul. 

TJie  snow  load  varies,  according  to  the  locality,  from  10  lbs. 
to  30  lbs.  per  square  foot  of  horizontal  projection.  The 
weight  of  new  snow  varies  from  <">  lbs.  to  12  lbs.  per  cubic 
foot,  although  snow  which  is  saturated  with  water  weighs 
nnich  more.  Tlie  snow  load  need  not  be  considered  when 
the  inclination  of  the  roof  to  the  horizontal  is  60°  or  more, 
as  the  snow  would  slide  off. 

The  luind  load  is  variable  in  direction  and  intensity,  and 
often  injurious  in  its  effects ;  especially  is  this  the  case  with 
large  trusses  pkced  at  considerable  intervals  apart.  The 
maximum  wincl  pressure  on  a  surface  normal  to  its  direction 
is  variously  estimated  at  from  30  lbs.  to  HG  lbs.  per  square 
foot.  The  calculation,  therefore,  of  the  stresses  cansed  by 
the  wind  forces  is  often  of  considerable  importance,  and 
should  never  be  left  out  of  account  in  designing  iron  roofs 
of  large  span. 

Art.  4.  The  Apex  Loads  and  Reactions.  —  Both  the 
dead  and  live  loads  are  transferred,  by  means  of  the  purlins, 
to  the  joints  or  apexes  of  the  upper  chords  of  the  truss ;  and 
these  loails  at  the  apexes  are  called  the  apex  loads.  They 
are  also  called  the  panel  loads. 

The  truss,  roof,  and  snow  loaiu,  being  vertical,  and  uni- 
formly distributed  over  each  panel,  the  apex  loads  are 
each  equal  to  one  half  the  sum  of  tl.e  adjacent  panel  loads. 
Thus,  the  load  at  h,  Fig.  1,  is  eoaal  to  one  half  the  panel 
load  on  Ac  plus  one  half  the  panel  load  on  ah.     The  wind 


•  . 


-',' 


ROOF  TliUSSES.  '•-.•'■> 

load  at  h  is  also  0(iiial  to  one  lialf  the  wind  load  on  Ac  plus 
one  half  the  wind  load  on  ah,  the  load  on  eaoh  being  normal 
to  the  surface. 

If  the  truss  bo  synnuetri- 
cal,  and  the  panels  be  of  equal 
length,  the  apex  loads  are 
d(;termined  by  dividing  the 
total  load  by  the  number  of 
panels  in  the  upper  chords. 
Thus  in  Fig.  1,  the  apex  loads  ^''*"  ^ 

at  h  and  c  are  each  one  fourth  of  the  total  load.  At  the  suii- 
ports,  a  and  l,  the  loads  are  only  one  half  those  at  h  and  c. 

Reactions.  —  For  dead  and  snow  loads  both  reactions  of 
the  supports  of  the  truss  are  vertical,  and  each  is  one  half 
of  the  total  load,  if  the  truss  is  symmetrical.  For  wind  load 
the  reactions  depend  upon  the  manner  of  supporting  the 

truss.  ■_  :    ■        ,; 

Pi'ob.  1.  A  roof  truss,  like  Fig.  1,  has  its  span  80  feet  and 
its  rise  40  feet;  the  distance  between  trusses  is  12  feet, 
center  to  center.  Find  (1)  the  weight  of  the  truss,  (2)  the 
weight  of  the  roof,  (3)  the  snow  load,  (4)  the  apex  loatls, 
and  (5)  the  reactions. 

For  these  problems,  take  20  lbs.  per  square  foot  of  roof  sin-face  for 
weiglit  of  roof,  and  20  lbs.  per  square  foot  of  horizontal  projection  for 
snow  load. 

Here  ab  =  80  feet,  dc  =  40  feet,  and  ac  =  5G.5C  feet. 

Weight  of  truss,  \V,  =  -,^br-  =    3200     lbs. 

Weight  of  roof,    TF,  r^  56.5G  x  20  x  12  x  2  =  27148.8  lbs. 
Weight  of  snow,  W,  =  80  x  12  x  20  =  19200     lbs. 

Each  apex  load  =  J  x  49.548.8  =  12.SS7.2  lbs. 

Each  reaction,       ii  =  J  x  49548.8  =  24774.4  lbs. 


i  ■'^'  ^  .. 


«-- 


>^ 


*-.- 


i   , 


*  ,. 


•*•." 


■MS&  ' 


i-:%*M  Mti  •aiwi'   '^^m 


6 


ROOFS  AND  li JUDGES. 


Proh.  2.    A  roof  truss,  like  Fig.  2,  has  its  si)au  00  feet,  its 
rise  ;«)  feet,  and  distance  between  trusses  la  feet:  find  the 


Via.  a  '■■  ,  "  ^'   :' 

total  apex  loads  and  the  reactions  for  the  weights  of  the 
truss,  roof,  and  snow. 

Ans.  Apex  loads  =  G989  and  13977  lbs.;  reaction  = 
27954  lbs. 

Pfob.  3.  In  the  roof  truss  of  Fig.  3,  the  span  is  100  feet, 
the  rise  is  25  feet,  and  the  distance  between  trusses  12.5 
feet:  find  the  total  apex  loads  and  the  reactions  for  the 
truss,  roof,  and  snow  loads. 


k  d 

FiK.  3 

Ans.  Loads  =  7270  and  14540  lbs. ;   reaction  =  29079  lbs. 

Art.  5.  Relations  between  External  Forces  and 
-^intemal  Stresses.  —  The  external  forces  acting  upon  a 
tvui^f-^j-fl  in  equilibrium  with  the  internal  stresses  in  the 
members'iS  r^^e  truss. 

Let  mn  be  a'stvi/^'n  passed  through  the  truss,  Fig.  4,  cut- 


#^ 


nnOF  TRUSSEfi. 


0  feet,  its 
find  the 


ts  of  the 
eaction  = 


1 100  feet, 
isses  12.5 
>s  for  the 


29079  lbs. 

ces  and 

ig  upon  a 
ies  in  the 

i'ig.  4,  cut- 


ting tlie  members  whcse  stresses  are  desired,  and  let  tliese 
stresses  be  replaced  by  equal  external  forces.     Then  it  is 


Fig.  4 


clear  that  the  equilibrium  is  undisturbed.  Therefore  we 
have  the  principle: 

The  internal  stresses  in  any  section  hold  in  equilibrium 
the  external  forces  acting  upon  either  side  of  that  section. 

If  we  remove  either  portion  of  the  truss,  as  the  one  on 
the  right  of  tlie  sectiovi,  then  the  external  forces  on  the 
remaining  part  of  the  truss,  together  with  the  internal 
stresses,  form  a  sy.stem  of  forces  in  static  equilibrium. 
And  from  Anal.  Mechanics,  Art.  Gl,  the  conditions  of 
equilibrium  for  a  system  of  forces  acting  in  any  direction 
in  one  plane  on  a  rigid  body  are: 

2  horizontal  components  =  0 (1) 

S  vertical  components      =0 (2) 

S  moments  =0 (3) 

These  three  equations  of  condition  state  the  relations 
between  the  internal  stresses  in  any  section,  and  the  ex- 
ternal forces  on  either  side  of  that  section.  If  only  three 
of  these  internal  stresf^es  are  unknown,  they  can  therefore 
be  determined. 

For  example,  in  Fig.  4,  let  R,  P^,  Pi  be  the  reaction  and 
apex  loads,  found  as  in  Art.  4;  let  .Sj,  .1.2, .%  be  the  stresses 
in  the  members  he,  kc,  and  kd,  that  are  (!ut  by  the  section 
mn,  and  let  fl,  and  63  be  the  angles  made  by  .s,  and  s^  with 


Jit' 


■'  <"- 


8 


fiooF.9  AND  ntiinnEsi. 


the  vertical.     Applying  our  throe  equations  of  equilibrium, 
wo  have : 

for  horizontal  components,  »,  sin  $i  +  a,  sin  Oj  +  Sg  —  0, 

for  vertical  components,  R - 1\— 1\  -\-  s,  cos  fl,  +  Sj  cos  0^=0. 

In  applying  our  third  equation  of  condition,  the  ct-nter  of 

moments  may  l)e  chosen  at  any  point,  Anal.  Mechanics, 

Art.  40.     If  we  take  it  at  c,  the  moments  of  s,  and  Sj,  are 

zero,  and  the  equation  is  — 

(R -  I\)\ ab-PiX  \ab - Sg  x  cd  =  0. 

These  three  equations  enable  us  to  find  the  unknown 
stresses.  ■'"■-  "  .  '  ;[.,^;'-  ■■■'^>;.  ' 

NoTK  1.  —In  all  cases,  in  this  work,  a  tensile  stress  is  denoted  by 
the  positive  sign,  and  a  compressive  stress,  by  the  negative  sign.*  In 
stating  the  equations,  it  will  bo  convenient  to  represent  the  unknown 
stresses  as  tensile,  pulling  away  from  the  section,  as  in  Fig.  4.  Then, 
if  the  numerical  values  of  these  stresse*  are  found  to  be  positive,  it 
will  show  that  tliey  were  iissuuied  in  the  right  direction,  and  are  ten- 
sile ;  but  if  they  are  negative,  they  were  assumed  in  tlie  wrong  direc- 
tion, and  are  compressive. 

Prob.  4.  In  the  truss  of  Fig.  5  the  span  is  90  feet,  the 
rise  is  35  feet,  hk  is  perpendicular  to  the  rafter  at  its  mid- 
point, and  the  loads  are  as  shown :  let  it  be  required  to  find 
the  stresses  in  all  the  members  of  the  truss. 

Representing  the  stresses  by  Si,  .%  s^,  S4,  Sj,  and  a^,  we 
proceed  to  apply  our  three  equations  of  equilibrium. 

To  find  the  stress  s„  pass  a  section  cutting  s,  and  Sg,  sepa- 
rating the  portion  to  the  left,  take  moments  about  h,  and 
regard  .<!,  as  pulling  toward  the  right  from  the  section 
(see  Note  1).     Then,  2  moments  about  h  =  0  gives 

(20000  -  5000)  22^  -  s,  x  .17^  =  0.     .-.  s,  =  19287  lbs. 

•This  convention  is  entirely  arliitrary.  Some  writers  denote  com- 
pression by  tlie  iwsitive  sign,  and  tension  l»y  the  negative  sign. 


i 


I 

1 


we 


I 


h 


HOOF  TnUSSES. 


To  fin.l  s,,  jKiHs  !i  section  cuttinj,'  s„  .^  un.l  «„  t,ak o- 

meats  about  c,  and  regard  «,  as  pulling  toward  the  right. 


.:   -A     I      '-^../    ^ 

■   toooo  ^,  toooo 

Via.  a 

S  mom.  about  c  =  0  gives 

(20000  -  5000)  45  -  10000  x  22J  -  /»,  x  35  =  0 

.-.  sa  =  12857  Iba. 
To  find  »„  pass  a  section  cutting  s,  and  »,;  we  might  take 
moments  about  k,  but  we  will  use  our  second  equation 
instead. 
2  vert.  comp.  =  0  gives  ■  :  { 

20000 -5000  + S3  sin  c«rf  =  0, 

15000  +  f « .,3  =  0  ("since  sin  cad  =  "^  =  ^^V 
V  ac     57) 

•••  »j  =  —  24428  lbs.,  that  is,  compression  (see  Note  1), 

To  find  s„  pass  a  section  cutting  s„  .9«,  and  s«  take  mo- 
ments about  k,  and  find  the  lever  arms. 
2  mom.  about  A;  =  0  gives 

(20000  -  5000)36.1  -  10000  x  13.6  +  «,  x  22.17  =  0 

(since  ak  =  ah  sec  cml  =  36.1  and  hk  =  ah  tan  cad  =  22.17). 

•■•  »<  =  v  18290  lbs.,  that  is,  compression. 


or, 


.1 


10 


HOOFS  AND   ItlllDdKS. 


To  fnid  .s,„  pass  thn  section  c.uUiiig  »„  a»,  and  s^,  and  tako 
nionicnts  al)oiit  n.     TImmi 

lOOOO  X  Y  +  «»  X  V  =  *^-     •■•  •"»  =  -  '''8i>5  lbs. 
To  liiiil  .Vfl,  pass  tln!  section  cuttinj,'  n.j,  .%,  and  s^,  and  take 
moments  about  a.     Tlion, 

10000  x22.5-8«x 35=0  (since  the  lever  arm  of  S8=c(«=35), 
.-.  .s„  =  (;  tL'.S  lbs. 
Since  the  truss  is  symmetrical  a.id  syuimetrieally  loaded, 
it  is  evident  that  the  stresses  in  all  the  pieces  of  the  right 
half  are  cMpial  to  those  just  found  in  the  left. 

NoTK  2.  —  In  tlio  ivbovo  Hulution  we  Imvo  called  forces  acting  up- 
wanls  iin<i  to  the  rif,'lit,  pimitivc,  iiiiil  forcuH  acting  ilownwarils  ami  to 
tilt!  left,  nei/dtivo ;  also  wo  have  t';illeil  iiKmient.s  ti'iidiiig  to  cause 
rotation  in  the  direction  of  tho  hands  of  a  clock  from  left  to  riKht, 
piisitivi>,  and  tluwe  in  the  oppn.sitc!  direction,  negative.  Tho  oiiposito 
coiivtuition  would  do  a«  well ;  we  have  only  to  introduce  opposite 
forces  and  opposite  nionients  with  unlike  HJgns. 

Prob.  5.  A  roof  truss  like  Fif,'.  1  has  its  span  40  feet,  its 
rise  20  feet,  and  the  apex  loads  2000  lbs. :  find  the  stresses 
in  ad,  ah,  he,  find  hd. 

Ans.  (id  =  +  1.5  tons,  ah  =  —  2.12  tcnis,  he  =  —  1.41  tons, 
hd  =  -  0.71  tons. 

Pri>b.  6.  A  truss  like  Fig.  2  has  its  span  (50  feet,  its  rise 
20  feet,  and  tlm  panel  loads  4000  lbs. :  find  the  stresses  in 
ad,  ah,  he,  and  hd. 

Aiis.  ad  =  4-  4.5  tons,  ah  =  —  5.40  tons,  /jo  =  —  3.G4  tons, 
hd  =  -  1.82  tons. 

Prnh.  7.  A  truss  like  Fig.  3  has  its  span  80  feet,  its  rise 
20  feet,  and  the  panel  load?  10,000  lbs. :  find  the  stresses 
in  all  the  memligrs. 

Ana.  ak  =  +  15  tons,  kd  =  +  10  tons,  ah  —  ~  10.75  tons, 
/«*=  — 1  1  5  tons,  /(A;=— 4.5  tons,  fcc= +5  trms,  rf?=00  tons. 


-t..  t,-iRre55wj^.:^'5«?tTS:'*«iE«^-ai 


g,J!!'!'yw-.t!l^'-' '    ■»T?ff^"' 


TnracBn^sacv.^  -11.-- 


ItoOF  TltUSSEfl. 


11 


Alt.  6.  Methods  of  Calculaf  on.  —  Our  tliipo  oq\ia- 
tioiis  III'  cipiililirimn,  Art.  r»,  I'liniisli  iis  witli  two  nit'lliiMis 
uf  ciilciil.'iliou  :  //(('  iiK'tliod  hi/  irniihtliiiii  nf  fiti'ci'n,  iiiul  tlif 
iiii'thdil  of  mniiu'iil.i.  'VUv  priiiciiik)  of  llin  lirst  im'tliod  is 
L'lubriiciul  in  tho  first  two  iMnuitioiiM  of  ('(luilil)iiuin ;  tlio 
Iiriiicii)U'  of  tlio  second  iiKitliod  is  enilnaced  in  tlie  third 
tM|niition  of  ('({iiililiriiini. 

Kitlun-  of  tiit'sc!  nu'tJKjd.s  may  be  used  in  tlie  solution  of 
ftny  KiviMi  case;  but  in  general  there  will  be  one,  the  em- 
jiloynn'Mt  of  whi(!h  in  an}'  sjiccial  ease;  will  bo  found  easier 
and  simpler  than  the  other.  Sometimes  a  eumbiimtion  of 
both  methods  furnishes  a  readier  solution. 


Kkmaiik.  —  A  section  may  bo  imsseil  tlir(m;,'li  iv  tni.ss  in  any  dirnc- 
ti(in,  sciianilinK  it  into  any  two  p'lvtions.  'Ilnis.  in  Fi^.  Ti,  a 
Hi'ctinn  may  In-  i)a.sHial  arouuil  /(,  cntlinj;  he.  hk\  and  hn.  'I'licii  llie 
inttTiial  Ntii'HHt'H  »i,  Hi,  «8,  and  the  apex  h)ad  of  10,(KX)  lbs.  at  h 
foi'in  ii  HyHtcui  (if  fiircus  in  t'(|ullibriuiii,  to  wliicii  our  ctiuntiouH  are 
applicabli'. 

A  judiciiiiis  .selieliiin  (if  dircctidns  for  tlie  ri'solutioii  of  tbc  forrcs 
often  Kimplities  Uic  dcti'rniinatiou  of  tlie  Hi rcasci.  Tluis,  to  iind 
.15  in  Fi^.  5,  if  wo  rcHolve  tlie  forces  into  a  direction  periiendicular 
to  tli(!  rafter  nr.  we  sliall  oblain  an  ei|nation  free  from  the  forces 
S:i  and  S4 ;  wliereius  if  the  directioiiH  are  talteii  at  randoni,  all  of 
tiie  forces  will  enter  the  O(ination.  This  principle  is  a  very  useful 
one. 

Thus,  we  have  at  onee  in  Fifj;.  /»,  ealling  0  the  angle 
between  the  load  and  the  rafter, 

10000  sin  e  +  .Si  =  0.      .-.  s,  =  -  10000  cos  cwl  =  -  7.S95  lbs. 

as  before. 

Similarly,  if  a  section  be  passed  around  d  in  Fig.  T),  cut- 
ting (Ik,  ik,  and  <Ui,  and  the  vertical  conqionents  be  taken, 
the  stress  in  </(•  is  fomul  at  onve.  to  be  zero. 

In  solving  by  the  second  method,  the  center  of  moments 


'^tefev 


i^^r 


1: 


HOOFS  A\it  nniDnFs. 


may  ho  takoii  anywlmro  in  tlip  plmie  nf  tlm  forcrs.  It  is 
idti'ii  iniiir  ntiivciiiciit  to  write  tliico  nioiiinit  (MiiiiitioiiH  lor 
tiio  stresMt's  in  tfin  tlnco  iiumiiImts  cut,  tiiiiiii^'  a  w\v  cciitfr 
of  iiiuiimnts  ciicli  tiiiif,  tiiiiii  it  is  to  uno  tlic  liist  two  (miiiii- 
tions  of  eriiiilil)riuiu;  and  if  tlio  center  of  nionit-nts  bo  talten 
at  the  intersection  of  two  of  tlio  pieces  cdt,  we  shall  have  at 
oneo  the  moment  of  the  stresH  in  the  other  jjiece,  balaneed 
by  the  sum  of  the  momentH  of  the  ext(irnal  forci  s,  since 
the  moments  of  the  stresses  in  tlie  other  two  ent  pieces  are 
zero. 

Tims,  in  solving  l'rol».  I,  a  section  was  passed  cutting,'  »j, 
«„,  St.  To  find  Aj  wo  took  jnomonts  about  c,  the  intersection 
of  »„  and  S4,  which  gave  us  itn  equation  of  nioiuents  con- 
taining  only  «,  anti  kiiown  terms.  To  find  s^  we  took 
moments  about  k,  the  intersection  of  h.^  and  %  whi(di  gave 
us  an  ecpiation  of  nuiments  containing  only  s^  and  known 
terms.  Also,  to  lind  n„  wo  took  moments  about  n,  the  inter- 
section of  n.i  and  a<,  whi(!h  gave  us  an  Cfpiation  of  moments 
containing  only  »«  and  known  terms. 

Therefore,  to  find  the  stress  in  any  member  by  the 
method  of  moments,  we  Iiavo  the  following  Jiule: 

Conceive  at  any  point  of  thin  member  n  nertion  to  he  pasned 
completcli/  thvDuijh  the  truss,  nUtinij  three  members.  To  fiinl 
the  sirens  in  this  member,  take  the  center  of  moments  at  the 
intersection  of  the  other  two.  Then  state  the  equation  of 
moments  between  this  stress  and  the  external  forces  on  the 
left  of  the  section. 

Should  the  section  that  passes  completely  through  the 
truss  cut  more  than  three  pieces  Avhose  stresses  are  un- 
known, if  Ail  but  that  piece  in  which  the  stress  is  reriuired 
meet  at  a  common  point,  the  center  of  moments  may  be 
taken  at  that  point. 


I 


V 

.% 


'<■ 


nooF  rnussics. 


SiKdild  the  HHition  cut  hut  two  pioees,  tho  center  of  mo 
lut'iits  for  oitlier  piei-o  may  be  tukcu  at  any  iioiut  of  the 
otlicr. 

I'nih  8.  Willi  ♦'it'  (UniensioUH  and  apex  loads  given  in 
Trob.  u  find  the  »tress  in  nl  of  Vig.  I,     (See  Item.) 

Ana.  -f-  1  ton. 

]\ofi,  9.  With  the  dimensions  and  apcv  loads  given  in 
Prob.  G  find  the  stress  in  cd  of  Kig.  li.  Aim  +  l.'J  tons. 

I'rnh.  10.  In  a  truss  like  Fig  4  the  span  is  80  feet, 
the  rise  of  truss  is  2('  feet,  the  rise  of  tie  rod  is  4  feet,  the 
panel  loads  are  cacli  4  tons:  find  the  stresses  in  all  the 
members. 

.  ]n.-t.  «A;  =  -f  18.4  tons,  kd  =  +  10  tons,  <ih  =  -  20.2  tons, 
he.  =  —  18.4  tons,  hk  —  —  3.G  tons,  kc  -■  +0  tons,  cd  =  0 
(only  siiitports  part  of  the  tie  rod). 

I'rob.  11.  In  Fig.  G  the  span  is  100  feet,  the  rise  of  truss 


■^ 


v\ 


£^K.  e 


is  2o  feet,  the  apex  loads  are  3  tons:  find  the  stresses  in  all 
the  members. 

Ana.  ah  =  -  16.8,  hi  =  -  13.44,  lc=  -  lO.OS,  ak=:+  Ifi, 
km=  +15,  OTd=  +12,  hk  =  0  (only  supports  part  of  the 
tie),  Im  =  -\  1.5,  cd  =  +  G,  hm  =  -  3.3G,  Id  =  ~  4.2. 

Suo. — The  stresses  iii  the  lower  chords  and  in  all  the  verticals 
except  the  center  one  are  best  found  by  the  method  of  inomunts ;  tlie 


I 


i 


14 


ROOFS  AND   niilDGES. 


stresses  in  the  upper  cliords,  and  also  in  tl\c  diajtonals,  may  be  fomul 
liy  I  lie  nielliod  by  resDliitlon  of  forcen,  althougli  Uie  strcHses  in  the 
diiij,'()nals  are  easily  found  by  tlie  nietliod  of  nionienls,  taking  tlie 
center  at  a.  To  find  tlio  stress  in  cd  it  is  best  to  pass  a  section  around 
d  and  take  vertical  components,  as  in  I'robs.  8  and  1». 

Prob.  12.  A  truss  like  Fig.  6  has  80  feet  span,  20  feet 
rise,  distance  between  trusses  12  feet,  and  Aveight  of  roof  20 
lbs.  per  s(]uare  foot  of  roof  surface  (see  l?i-ob.  1) :  tind  the 
dead  load  stresses  in  all  the  members,  in  tons. 

Ans.  ah  =  -11.5,  Jil  =  -d.'2l,  fc  =  -6.9,  aA;=: +10.28  = 
km,  mil  = +8.22,  hk  =  0,  Zwi=+1.03,  cd=+4.11,  hm  = 
-2.3,  til  =-2.88. 

Art.  7.    Lever  Arum  —  Indeterminate  Cases.  —  In 

determining  the  stresses  by  the  method  of  moments  the 
only  difficulty  lies  in  finding  the  laver  arms  of  the  vaiious 
pieces.  These  can  always  be  found  by  the  use  of  geometry 
and  trigonometry.  Thus,  in  Fig.  G,  the  lever  arms  for  the 
lower  panels  are  evidently  the  perpendiculars  let  fall  upon 
these  panel?  from  each  opposite  upper  apex.  The  lever 
arms  for  the  upper  panels  are  the  perpendiculars  drawn  to 
these  panels  from  each  opposite  lower  apex.  The  lever 
arm  for  eacli  brace  is  the  perpendicidar  to  the  direction  of 
the  brace  drawn  from  the  left  end  a  of  the  truss,  v/liere  the 
rafter  and  tie  intersect.  This  is  evident  from  our  rule  in 
Art.  6. 

Thus,  in  Fig.  6,  suppose  a  section  to  cut  hi,  hm,  and  km. 
By  our  rule  the  center  of  moments  for  km  is  h,  the  point  of 
intersection  of  the  other  two  pieces  "  I  and  hm.  For  hi  the 
center  is  m,  the  intersection  of  hm  und  km.  For  hm  the 
center  is  a,  the  intersection  of  hi  and  km. 

For  trusses  in  which  the  members  h  we  various  inclina- 


HOOF  TRUSSES. 

tions,  all  tliffei-ent,  the  computatiou  of  the  lever  arms  is 
quite  tedious.  In  such  cases,  it  is  soiuetiuies  ailvisable  to 
make  a  careful  drawiug  of  the  truss,  and  tlu-n  measure  the 
lever  arms  by  scale.  Indeed,  this  method  can,  in  all  cases, 
be  used  as  a  check  upon  the  accuracy  of  the  results  obtained 
for  the  lever  arm  by  computation. 

If  the  section  dividing  the  truss  into  two  parts  cuts 
more  than  three  members,  the  stresses  in  which  are  un- 
known, the  problem  is  indeterminate,  because  there  are  more 
unknown  quantities  to  be  found  than  there  are  Cijuations  of 
condition  between  them.  In  such  cases  a  fourth  condi- 
tion is  sometimes  found  in  the  symmetry  of  the  truss  and 
loads. 

Thus,  if  we  pass  a  section  through  cl,  cd,  dV,  and  dm', 
Fig.  (5,  it  cuts  more  than  three  pieces.  l?ut  the  stress  in  dV 
is  equal  to  that  ii\  dl,  by  reason  of  the  symmetry  of  the 
truss  and  loads.  Even  if  this  were  not  the  case,  it  could 
easily  be  found  by  working  toward  it  from  the  right  end. 
Then  there  remain  only  three  unknown  quantities,  whose 
values  can  be  determined  by  our  three  equations. 

The  section  may  cut  any  number  of  members,  so  long  as 
it  is  possible  to  find  independently  the  stresses  in  all  but 
three.  Any  truss  which  violates  this  rule  is  improperly 
framed,  and  has  wmecessmij  or  .mper/lnous  pieces. 


liEMAKK.  — The  half  of  each  end  panel  load  carried  by  the  support 
floes  not  affect  the  truss,  and  need  not  be  taken  into  account  in  find- 
ing cither  ioads  or  reactions.  Thus,  in  Fij;.  5,  tlie  reaction  and  tiie 
half  panel  load  acting  at  the  support  are  ecjuivaleiU  to  an  upward 
force  equal  to  then-  difference.  This  upward  force  is  the  reaction 
found  by  c.initting  the  two  apex  loads  at  the  supports,  and  is  known 
a.s  the  effective  or  workinn  reaction.  Thus,  instead  of  calling  20000 
the  reaction,  and  writing  the  equation 


20000  X  22*  -  5000  x  22- 


Si  X  17'  =  0, 


1-: 


10 


ROOFS  AND  BUI  DOES. 


us  is  done  in  the  soluUnii  of  Prob.  4,  we  call  i!iu  reaction  15000  and 
write  the  e<iuatiou  as  follow-- 

15000  X  22-  -si  X  17*  =0, 
and  so  for  all  the  other  nwiiient  equations. 

I'rob.  13.   A  truss  like  Fig.  7  has  80  feet  span,  20  feet  rise, 
distance  betweeu  trusses  12  feet,  and  weight  of  roof  aa 


S.S5 


before  (see  Trob.  1) :  find  the  dead  load  stresses  in  all  the 
nieiuhers. 
Here  the  dead  apex  load 

12  X  Wf 


■f  12  X  2a  V5«+ 10'  =  3083  lbs. 


24  X  8 
=  1.54  tons,  say  1.5  tons. 

Effective  reaction  =  1.5  x  3.5  =  5.25  tons  (see  remark). 
To  liud  fy  =  af,  take  inonients  around  b. 

5.25  X  10  -/j/  X  5  =  0.    .-.  fg=+ 10.6  tona.   , 

To  find  yh  take  moments  around  c. 

5.25  X  20 -  1.5  X  10 - ///i  x  10  =  0.     .:  gh=+9  tons. 

To  find  ijr-pass  section  around  g  (see  rem.  of  Art.  G). 


10.5  +  bgx 


10 
11.18 


9  =  0.     .-.  by  =—1.68  tons;  and  soon. 


t 


J. 


i- 


V^.C^.'.t-t  --XtHitsiSaimiii^s^KilXF--^- 


n  15000  asid 


!0  feet  risp, 
of  roof  as 


in  all  the 


)S. 

Binark). 
ns. 

-  9  tons, 
rt.  G). 

and  so  on. 


ROOF  TliUSSES. 


Ans.  ah  =  - 11 .76,  he  =  -  10.08,  cd=  -  8.4,  dc-  -  6.72, 
a/=  + 10.5,  /£/  =  + 10.5,  (jh  =  -I-  9,  hk  =  +  7.5,  />/=  0  (only 
supports  part  of  the  tie  rod),  cflf  =  +  0.75,  ah  =  +  1.5, 
ck=+  4.5,  bg  =  -  1.68,  e/t  =  -  2.15,  dk=-  2.7. 

Prob.  14.  In  Fig.  8  the  span  's  60  feet,  the  rise  of  truss 
is  15  feet,  the  rise  of  tie  rod  at  point  ^  is  2  feet,  the  panel 


Viit. 


loads  are  1.6  tons,  the  struts  3-4,  5-6,  7-8  are  vertical, 
dividing  the  rafter  2-7  into  three  equal  parts:  find  the 
stresses  in  all  the  members. 

Ans.  Stress  in  2-3  =  -  9.75,  in  3-6  =  ~  7.8,  in  5-7  = 
-5.85,  in  2-4=4-8.73,  in  4-6  =  +  8.73,  in  6-8  =  +  7,  in 
3-4  =  0  (only  supports  part  of  tie  rod),  in  5-6  =  -|-  0.75, 
in  7-8  =  -H  3.7,  in  3-6  =  - 1.85,  in  5-«  =  -  2.24. 

Proh.  16.  In  Fig.  9  the  span  is  90  feet,  the  rise  of  truss  is 
22.6  fe«t,  the  rise  of  tie  rod  is  3  feet,  the  struts,  3-4,  5-6, 


I'Ms.  O 


7-8,  and  9-10  are  vertical,  dividing  the  rafter  into  four 
ecpial  parts;  the  apex  loads  are  2  tons:  find  the  stresses  in 
all  the  members. 


18 


HOOFS   A  Nit   llliinaEN. 


Aiix.  Sticbs  in  2-3  =  - 18,  in  .'{-5  =  -  15.4,  in  5-7  --=  - 
lli.S,  in  7-1)  =  -  10.2,  in  2-1  =  +  IG.l,  in  4-(}  =  +  U).l,  in 
(•,-8  =  +  13.8,  in  8-10-r  4- ll.u,  in  3-4^=0  (this  is  not 
necessary  to  the  staliility  of  tlie  truss),  in  5-()  =  +  1.0,  in 
7-8  =+2.0,  in  9-10  =+7.2,  in  3-6  =-2.46,  in  5-8  = 
-  2.96,  in  7-10  =  •-  3.68. 

Pmb.  16.  In  Fig.  10  the  span  is  120  fc't,  the  rise  is  30 
feet,  the  struts  3-4,  6-6,  7-8  are  drawn  normal   to  the 


! 


rafter,  dividing  it  into  four  equal  parts,  the  apex  loads  are 
2.5  tons :  find  the  stresses  in  all  the  members. 

Atis.  Stress  in  2-3  =  -  19.5,  3-5  =  -  18.38,  5-7  = 
-  17.25,  7-9  =  -  16.13,  2-4  =  +  17.5, 4-6  =  +  15.0, 6-10  = 
+  10.0,  3-A  =  -  2.23,  5-6=  -4.45,  7-8=  -2.23;  4-5=  +2.5, 
5-8  =  +  2.5,  6-8  =  +  5.0,  8-9  =  +  7.5,  9-10  =  0  (only  sup- 
ports part  of  the  tie  rod,  and  not  necessary  to  the  stability 
of  the  truss). 

It  will  be  observed  fhat  the  members  3-4  and  7-8  are  symmetrical 
with  respect  to  their  loads,  and  therefore  their  stresses  arc  equal,  and 
also  that  5-4  and  6-8  are  symmetrical,  and  hence  their  stresses  are 
equal. 

Prob.  17.  In  Fig.  11  the  span  is  90  feet,  the  rise  of  truss 
is  22.5  feet,  the  rise  of  tie  rod  is  3  feet,  the  struts  3—4,  5-6, 
7-8  are  drawn  normal  to  the  rafter,  dividing  it  into  four 
equal  parts ;  the  apex  loads  are  2  tons ;  lind  the  stresses  in 
all  the  members.  • 


4k 


■?f- 


in  r»-7  .-=  - 
=  4-  Ki.l,  in 
(this  is  not 
I  =  +  1.0,  in 
6,   in  5-8  = 

he  rise  is  30 
riual   to   the 


pex  loads  are 

8.38,  5-7  = 
15.0,  G-10  = 
;  4-5= +2.5, 
;  0  (only  sup- 
the  stability 

re  symmetrical 
.  arc  eciual,  and 
eir  sti-esses  are 

I  rise  of  truss 

•II  ts  3-4,  5-6, 

it  into  four 

lie  stresses  in 


ROOF  TIIUSSES. 


Ans.  Stress  in  2-3=  -20.34,  3-n=  - 19.44,  5-7=  -18.54, 
7-9  =  --17.(J4,  2-4  =  +  18.3,  4-0  =  +  lo.GH,  0-10  =  +  9.3, 
3^-4  =  - 1.78,  5-0  =  -3.5(;,  7-8  =  -1.78,  4-5  =  + 2.64, 
5-8  =  +  2.64,  6-8  =  +  ().84,  8-9  =  +  9.48,  9-10  =  0  (not 
necessary  to  stability  of  truss). 


Art  8.  Snow  Load  Stresses.  —  The  snow  load  is 
estimated  per  stpiare  foot  of  horizontal  projection  (Art.  3). 
If  the  main  rafters  of  the  truss  are  straight  —  as  in  all  of 
our  previous  publems  — the  snow  load  is  uniformly  dis- 
tributed over  the  whole  roof,  and  the  apex  paow  loads  are 
all  equal  (see  Art.  4);  therefore  tlie  dead  load  and  snow 
load  stresses  are  proportional  to  the  corresponding  apex 
loads. 

Thus,  in  Prob.  13,  the  dead  panel  load  was  3083  lbs.  and 
the  snow  panel  load  =  10  x  20  x  12  =  2400  lbs.  Therefore 
\2  we  multiply  each  dead  load  stress  by  ^^§^3  (=.778),  we 
shall  have  the  corresponding  snow  load  stress. 

IJut  if  the  main  rafters  are  not  straight,  the  snow  load 
is  not  uniformly  distributed  over  tlie  whole  roof,  and  the 
apex  snow  loads  are  not  all  equal;  in  such  case,  the  snow 
load  stresses  have  to  be  determined  independently. 

Thus,  with  the  dimensions  given  in  Fig.  12,  for  9  feet 
between  trusses,  we  have  the  apex  snow  load  at  b  and  at  6' 
each  equal  to  one  half  the  ]»anel  load  on  ub  plus  one  half 
the  panel  load  on  6c         ' 

=  6  X  20  X  9  =  1080  lbs.  =  0.54  ton, 


'-■ «"'g.ji3;r" 


m 


20 


nOOFH  AND  nUIDGES. 


The  apex  snow  load  at  c  equals  one  half  the  panel  load 
on  cb  plus  one  half  the  panel  loiui  on  cb' 

=  8  X  20  X  9  =  1440  lbs.  =  0.72  ton. 
The  stresses  duo  to  these  apex  snow  loads  may  now  be 
found  in  the  same  way  as  the  deail  load  stresses  were  found 
in  the  preceding  problems. 


Via.  I'-a 

It  is  possible  for  one  side  only  of  a  roof  to  be  loaded  with 
snow.  This  possibility  is  recognized  in  desi"  ing  roofs  of 
very  large  span,  such  as  the  roof  of  the  Jersey  City  train 
shed  of  the  Pennsylvania  R.R.  In  special  forms  of  trusses 
such  a  distribution  of  snow  load  may  produce  a  maximum 
stress  in  some  members. 

Problem.  With  the  dimensions  and  apex  snow  loads 
above  given,  find  the  snow  load  stresses  in  all  the  members 
of  Fig.  12. 

Am.  ui'*  =  -1.44,  be  =  -1.3,  oti  =  + 0.86,  W  =  +  0.68, 
dh  =  +  i:.i,  <k  =  +  i).Oii. 

Art.  9.  Wind  Loads.  —  One  of  the  most  important 
(piestions  to  be  dealt  with  in  the  construction  of  roof-trusses 
is  the  pressure  of  the  wind;  for  it  is  evident  that  the 
stability  of  suc^h  a  structure  depends  upon  its  power  of 
carrying  not  only  the  weight  of  the  truss,  roof,  snow,  etc., 
but  also  the  pressure  caused  by  the  wind.  In  the  case  of 
large  trusses,  it  will  often  be  found  th-^t,  notwithstanding 
the  great  weight  of  the  structure,  the  stresses  produced  in 
some  of  the  members  by  a  gale  of  wind  are  almost  or  quite 


",l--Jltmv»^.VIkir'M''Ai!i:&t.»kM'mk. 


panel  lomX 


!iy  now  be 
were  found 


loaded  with 
ing  roofs  of 
r  City  train 
13  of  trusses 
a  maximum 

snow  loads 
;he  members 

M  =  +  0.58, 

t  important 
'  roof-trusses 
nt  that  the 
ts  power  of 
f,  snow,  etc., 
I  the  case  of 
withstanding 
produced  in 
lost  or  quite 


ROOF  thussks. 


•21 


as  f^reat  as  those  prodviced  in  the  same  pieces  by  the  dcail 
load.  It  appears,  therefore,  that  in  desij^iiinj,'  roof.s,  especially 
iron  roofs  of  large  span,  a  correct  estimate  of  tlie  wind  loads 
is  quite  as  important  as  a  correct  estimate  of  the  dead  loads. 

And  yet,  the  subject  of  wind  forces  is  not  well  under- 
stood. It  is  hardly  possible  to  define,  with  any  precision, 
what  degree  of  violence  can  lie  taken  to  represent  the 
greatest  wind  storm  that  has  to  be  provided  against,  in 
estimating  the  wind  pressure  anil  the  resulting  stresses  in 
the  members  of  a  truss,  the  practice  of  engineers  has  varied 
greatly.  Until  recently,  it  has  been  considered  in  England 
sufficient  to  provide  for  a  wind  force  of  30  to  40  lbs.  per 
•  square  foot  of  surface  normal  to  its  direction ;  while  in  this 
country  the  figures  have  been  taken  at  30  to  flO  lbs.  per 
square  foot. 

Taking  the  maximum  wind  pressure  against  a  surface 
normal  to  its  dir«cti;(n  as  50  lbs.  per  square  foot,  we  shall 
probably  be  on  the  side  of  safety. 

The  following  table  gives  the  normal  wind  pressure  per 
square  foot  in  pounds  for  different  inclinjitions  of  the  roof 
equivalent  to  a  horizontal  wind  pressure  of  50  lbs.  per 
square  foot,  calculated  by  Hutton's  formula: 


Imclin. 

Nor.  Pb»i. 

Inpi.in. 

NOK.  VHt» 

10° 

12.1  , 

33°  30' 

(i  span)  36.6 

16° 

18.0 

35° 

37.8 

20° 

22.6 

40° 

41.6 

21°  48' 

(J  span)  25.2 

46° 

43.0 

25° 

28.8 

50° 

47.6 

26°  34' 

(J  span)  :W.2 

66° 

41).  6 

30° 

33.0 

60° 

50.0 

For  inclinations  greater  than  60°  the  normal  pressure  per 
square  foot  is  50  lbs.  For  intermediate  inclinations  we  can 
fiad  the  pressures  by  iuterpolation. 


i  Is 


'I 

if 


I      i 


s^s;?- 


"'i'lgJKiJT 


' 


ItOOFS  AND  n RIDGES. 

Art.  10.  Wind  Apex  Loads  and  Reactions,  r.ct; 
Fig.  13  represent  a  roof  truss,  its  sjnin  Ix'in;^'  1(10  IVit.  and 
its  rise  '2'>  feet;  let  tlie  liistiuu'e  between  trusses  be  lli  •eet, 
and  suppose  the  wind  to  be  on  the  left  side.  Then  the 
/'tclination  of  the  rafter  ae  to  the  horizon  =  tair' J=L'«;".'M'. 
Heuw,  ?>i>iu  our  table,  the  normal  wind  pressure  per  s(piaru 


Vie.  13 


foot  =  .30.2  lbs.  The  total  normal  wind  pressure  on  the 
si-'s  of  the  roof  ae  is  therefore 

=  30.2  X  12  X  Vfi(F+W=  30.2  X  12  X  56.9  =  20258  lbs., 

one  fourth  of  which,  or  5004.5  lbs.,  is  the  pressure  on  cacli 
panel.  Jlenee  the  normal  wind  load  at  each  apex  b,  e,  d,  is 
5004.5  lbs.,  or  say,  in  round  numbers,  5000  lbs.,  or  2.5  tons, 
and  at  each  apex  a,  and  <>,  it  is  2500  lbs.,  or  1.25  tons 
(Art.  4). 

The  Reactions  caused  by  the  wind  pressure  are  inclined; 
the  horizontal  component  of  the  wind  tends  to  slide  the 
entire  truss  ot?  its  supports.  The  weight  of  the  truss  and 
the  roof  are  usually  sufficient  to  cause  friction  enough  to 
hold  it  in  place.  But  if  it  is  necessary,  the  truss  should  be 
fastened  at  its  ends  to  the  wall.  Roof  trusses  of  short  span, 
and  especially  wooden  trusses,  have  generally  both  ends 
fixed  to  the  supporting  walls.  But  la'-ge  iron  trusses  have 
only  one  end  fixed,  while  the  other  end  is  free,  and  resting 
upon  friction  rollers,  so  that  it  may  move  horizontally, 


7i-^m-'v!^!mmmmk 


tions.     r.ft 

110  I'.'ft.  ami 
^  be  lli  •Vt't, 
Tlien  the 
■' J  =!-'(;"  .'M'. 
5  per  t)(|tiiiiu 


lure  on  the 

J02o8  lbs., 

lire  on  eacli 
3x  b,  e,  (I,  is 
or  2.0  tons, 
V  1.25  tons 

re  inclined; 

0  slide  the 
i  truss  and 

1  enough  to 
s  should  be 
short  span, 

both  ends 
I'lisses  have 
md  resting 
urizontally, 


nOOF  TRUSSES. 

under  changes  of  teniporaturo.  Wo  have  then  two  cases: 
till!  first,  when  both  ends  are  fixed;  the  .scm  oiid,  when  one 
end  only  is  fixed,  and  the  other  end  is  free  to  move  upon 
rollers. 

Case  I.  When  both  ends  are  /.mi.  — Let  Fig.  13  represent 
a  roof  truss  with  both  ends  fixed,  its  span  being  100  feet, 
its  rise  25  feet,  and  the  wind  ajicx  loads  1.2r>,  2.5,  2.5,  2.5, 
and  1.25  tons,  as  found  above.  In  this  case,  the  two  reac- 
tions Ri  and  ii,  are  jiarallel  to  the  normal  wind  loads,  and 
may  easily  be  found,  as  follows : 

Let  6  be  the  angle  between  the  rafter  and  the  lower 
ohoid  ak,  and  take  moments  about  the  left  end  a.    We 

have  then 

< 

i?j  X  ah  -  (?  5  X  ab  +  2.5  x  ac  +  2.5  x  ail  + 1.25  x  ae)  =  0, 
or,  we  may  take  the  ic.'jidtant  of  all  the  loads,  or  10  tons, 

acting  r.t  c, 

.-.  R,  X  100  cos  ^  -  10  X  27.05  =  0, 

.-.  Rt  =  3.13  tons  (since  cos  0  =  ^S). 

The  reaction  Ri  may  be  found  by  subtracting  T?,  from 
the  total  wind  load,  giving  us  /i,  =  6.87  tons.  Or,  we  may 
find  Ri  by  taking  moments  about  the  right  end  k.    Thus, 

RiXah  —  10  (ah  —  ac)  =  0, 

from  which  we  get  Ri  —  6.87  tons,  as  before. 

Case  II.  When  one  end  is  fixed  and  the  other  is  free.  — 
(1)  Suppose  the  right  end  of  the  truss  to  be  free,  and  the 
wind  blowing  on  the  fixed  side,  as  in  Fig.  14.  The  right  end 
of  the  truss  is  supposed  to  rest  upon  rollers,  the  support  at 
o  taking  all  the  horizontal  thrust  due  to  the  wind.  This 
being  the  case,  the  reaction  iij  at  the  free  end  will  be  verti- 
cal, and  the  reaction  at  the  fixed  end  will  be  inclined. 


#» 


24 


HOOFS  AND  II  It  I  DOBS. 


'A 


y 


T(i  find  R.2  take  nionunts  about  n  ;  let  the  (liinciisioiiR  nii.i 
wind  loads  ho  the  siiiiic  as  in  Ki;;.  1.'5.     Thus, 

Jij  X  100  -  10  X  27.05  =  0.        .-.  Ji^  =  2.8  toua. 


♦  11/       II       n      J       71      0       /■      4 
•B.  .  Ib. 

Klg.  14 

Resolve  the  left  reaction  into  its  horizontal  and  vertical 
components,  //  and  /?„  thus, 

2  hor.  comp.  =  0 
gives  10  8infl-//=0.        .-.  H=iA6. 

S  ver.  comp.  =  0 

gives        -B, -I- 2.8 -10  cos  ^  =  0.        .-.  /?,  =  6.13. 

Or,  we  might  find  72,  by  taking  moments  about  the  right 
support;  thus, 

/?,  X  100  -  10 (100  cos  e  -  28)  =  0.     .:  Ri  =  6.13,  as  before. 

(2)  Suppose  the  wind  to  blow  on  the  free  side  ek  of  the 
truss,  Fig.  14.  The  reaction  iJ,  is  vertical,  as  before,  and 
the  reaction  at  the  fixed  end  a  may  be  resolved  into  its 
horizontal  and  vertical  components,  in  the  same  way  as 
above.     Thus,  we  find, 

//=4.'G,     i?,  =  2.8,    i2,  =  6.13; 

that  is,  when  the  wind  changes  from  one  side  of  a  roof  to 
the  other,  the  horizontal  component  //  has  the  same  value 
as  before,  but  acts  in  the  opposite  direction,  and  the  re- 
actions Ri  and  li^  interchange  their  values. 


rm 


tuut). 


nd  vertical 


3. 

■  the  right 

,  as  before. 

ek  of  the 
)efore,  and 
)d  into  its 
le  way  as 


a  roof  to 
ame  value 
id  the  re- 


mbtmumm 


ROOF  TRVSSKS. 


86 


rri>h.  18.  A  trims  like  Fi^,'.  14  lias  its  span  40  feet,  rise 
Kl  IVcl.  mill  t\u'  lotul  noi'iiial  wind  load  on  tin*  Hxed  side  li.'J 
tons:  liiid  tlio  wind  load  reactions. 

Ann.  /;,  =  1.90,  Itt  =  0.90,  //=  1.43  tons. 

Proh.  10.  A  truss  like  Fi>j.  3  has  its  span  fiO  feet,  its  rise 
12.5  feet,  and  tlu'  distance  between  the  trusses  8  feet:  find 
the  reactions  when  both  e- :L  are  fixed. 

Ana.  /f,  =  2.32,  /i,  =  l.Ofi  tons. 

Prob.  20.  A  truss  like  Fif?.  4,  with  one  end  free,  has  its 
t.Mi  80  feet,  it,s  rise  20  feet,  and  the  total  normal  wind  load 
on  the  fixed  sic'ic  5  tons  :  find  the  wind  load  reactions. 

Au8.  R^  =  3.07,  /;,  =  1.4,  //=  2.23  tons. 

Proh.  21.  A  truss  like  Fig.  4,  with  one  end  free,  has  its 
span  (50  feet,  rise  of  truss  12  feet,  rise  of  tie  rod  2  feet,  and 
the  total  normal  wind  load  on  the  fixed  side  4.5  tons :  find 
the  wind  load  reactions. 

Ana.  Hi  =  2.97,  li,  =  1.21,  H=  1.67  tons. 

Prob.  22.  A  truss  like  Fig.  6,  with  one  end  free,  has  its 
span  90  feet,  its  rise  18  feet,  and  the  total  normal  wind  load 
on  the  fixed  side  6  tons :  find  the  wind  load  reactions. 

Ana.  Ill  =  3.94,  R^  =  1.62,  //=  2.25  tons. 

Art.  11.  Wind  Stresses.  —  (1)  If  the  truss  have  both 
ends  fixed,  we  must  consider  the  wind  blowing  normally  to 
the  principal  rafters  on  one  side  only  (either  side  indiffer- 
ently) ;  and  as  the  wind  load  is  unsymmetrical  to  the  roof, 
the  wind  stresses  in  the  members  oi.  one  side  of  the  truss 
are  different  >om  those  in  the  corresponding  members  on 
the  other  side,  and  hence  they  must  be  computed  for  every 
member  in  the  truss. 

(2)   In  trusses  with  one  end  fixed  and  the  other  free,  we 


I 


J^ 


r 


» 


'^ 


I 


26  IKxn-S  AM)   It II  11)11  KS. 

uwut  coiisii!  T  Mio  wind  lilo'viiii,'  lirHt  on  onn  siiln  of  tho 
ti'iiHS,  and  lilt  <  on  tin-  (itlicr;  and  llii'  Ht.ri'.s.scs  inodiu'cd  in 
tlio  twr)  casi's  will  liiivn  to  ho,  (jomimti'd. 

Prof).  23.  A  truss  liko  Fig.  15,  with  both  ends  fixed,  has 
its  span  HO  feet,  its  riso  12,5  feet,  and  tho  wind  loads  and 
reactions  as  sliown:  lind  all  the  wind  stresses. 


Wo  find  «c  =  27.95  feet,  or,  in  round  numbers,  28  feet : 
ak  =  15.()S  feet  =  kc ;  hk  =  l  feet. 

Ilopresentiiif^  the  stresses  by  «„  Sj,  Sj,  s^,  Sj,  and  s^,  and 
applying  the  printuples  of  Arts.  5  and  G,  we  have  for  the 
left  half  of  the  truss : 


2.(52  X  14  -  .s,  X  fi.25  =  0, 

2.G2  X  28  -  3  X  14  -  h.,  x  12.5  =  0, 

2.62  X  14  +  s,,  X  7       =0, 

3  +  »,  =0, 

-  3  X  14  +  «8  X  12.5  =  0, 


,s,  =  4-  5.87  tons. 
Sj,  =  +  2.51  tons. 
Sg  =  —  5.24  tons  =  s^. 
sj  =  -  3.00  tons. 
Sfl  =  +  3.36  tons. 


For  the  right  half  it  is  better  to  resolve  the  right  hand 
rcaetion  1.88  into  its  horizontal  and  vertical  components, 

thus, 

//=  1.88  sin  e  =  0.84,  and   V  =  1.88  cos  d  =  1 .68  tons, 

and  to  state  the  e(piation  of  each  piece  including  the  ex- 
ternal forces  on  the  right  of  the  section  rather  than  on  the 
left.     Thus, 


'  ■•'i;iJii51!i>i*Sniym»JJ-'!MW«'at*ll!alte-  ■ 


tmmmtm 


ItnoF  TitirssEs, 


27 


I  4 


i.OSx  t2.n-.HI  x(5.1»r.  -Vxr..2n=:(),  .-.  h,'^  fli.r.2  tons. 

i.(Wx2r.-.si  X  \'j.r>^.s,'x\'j.r>^(),  .-.  «;=4--w  mns. 

l.«W  X  1  <-),(;«  f.v,,'  X  7^0,  .-.  *,'=  -.'{  "(>  tons -:=«;. 

I'nih.  24.  A  truss  likn  Fi(,'.  lo,  with  oiio  end  frco,  liiis  its 
span  40  leot,  its  riso  10  feet,  ami  tlio  total  normal  wind  load 
on  till!  lixcd  siilo  .'{.U  tons;  tind  all  tlio  wind  stresses. 

Wo  must  lirst  find  tho  reactions  and  horizontal  eonipo- 
nent,  as  in  Case  II.    Thus, 

/?,  =  1.90,     Ji,  =  0.9,     //  =  1.43. 

Ans.  A<,  =  +  ;{.«,  »,=  +  1.8,  /»a=:-2.8,  «4=-2.8,  s,= -1.6, 
a„=  +  i.8,  .V=+1.8,  s;=-2,  ,V=-2,  ^=0,  «„'=0. 

8i  o.  —  It  will  often  bo  best  to  state  tlib  equation,  uhIiik  the  ex- 
ternal forces  on  the  riylU  of  the  section.  Thus,  to  C'.nl  tlii'  stress  In 
«2-     It'  wo  use  tho  forces  on  the  right  of  the  section,  the  eciuation  is 

.Ox20-»aX  10  =  0.         .-.  »2  =  1.H. 

Hut  If  wo  use  tho  forces  on  tho  left  of  the  section,  iho  eijuallon  la 

1.00  X  20  +  1.4.}  X  10-32  x  11.18 -»j  x  10  =  0.       .-.  «,  =  1.8. 

Of  course,  to  find  tho  8tres.ses  in  members  near  the  loft  end,  not  so 
much  is  gained  by  using  tho  forces  on  tho  right  of  the  section. 

Prob.  26.  A  truss  like  l*'ig.  4,  with  one  end  free,  has  its 
span  (50  feet,  its  rise  lit  feet,  rise  of  tie  rod  .3  feet,  and  the 
total  normal  wind  loud  on  the  fixed  side  .'i  ton  :  tind  the 
wind  stresses  in  all  the  members. 

Ana.  Stress  in  ak  =  +  S.7,  kd  =  + 3.5,  «/t  =  -7.78,  he 
=  -  7.78,  hk  =  -  2."),  kc  =  +  r>Ar>,  bk'  =  +  4..3(i,  bh'  =  -  4.8, 
h'c  =  -  4.8,  h'k'  =  0,  ck'  =  +  1.0,  cd  =  0  (n(»t  necessary  to 
stability  of  structure). 

Prob.  26.  A  truss  like  Fig.  14,  with  one  end  free,  has  its 
span  80  feet,  its  rise  20  feet,  and  the  total  normal  wind  load 


>f  ^^1  .wiKimni 


■^Tvr-r 


•^•SC~-f**"**- 


># 


ill 


28 


ROOFS  AND  nUWGES. 


on  the  fixed  side  8  tons :  find  the  wind  stresses  in  all  the 
members 

Ans.  Stress  in  «/=  + 11.16  =yj/,  gh  =  +  S.92,  /(j  =  +  G.()8, 
bg  =  -  2.5,  c/t  =  -  3.10,  dj  =  -4.02,  eg  =  +1.12,  dh  =  +  2.24, 
e;  =  +  3.35,  ab  =  -9,  be  =  -7.5,  cd  =  -(j.O,  tie  =  -4.5, 
kf  =:f'g'  =  g'h'  =  h'J  =  +  4.47,  Vg'  =  e'h'  =  d'j  =  0,  b'f  =  e'g' 
=  d'h'  =  0,  kb'  =  b'c'  =  e'C  =  d'e  =  -  5.0  tons. 

Prob.  27.  A  truss  like  Fig.  16,  with  one  end  free,  has  its 
span  60  feet,  its  rise  12  feet,  the  struts  hk  and  h'k'  normal 


to  the  rafters  at  their  middle  points,  and  the  total  normal 
wind  load  on  the  fixed  sida  ac  4  tons:  find  the  wind 
stresses  in  all  the  members. 

Ans.  Stress  in  ak  =  +  5.4,  kd  =  +  2.7,  ah  =  —  4.6  =  he, 
hk  =  -2,  cfc  =  +  2.7,  bk' =  + 2.7  =  dk',  bh' =  -2.91  =  ch', 
h'k'  =  0  =  ek'. 

Prob.  28.  In  the  same  truss  Fig.  16,  with  the  same  dimen- 
sions as  given  in  Prob.  27,  let  the  same  normal  wind  load  of 
4  tons  blow  on  the  free  side  be:  find  the  wind  stresses  in 
all  the  members. 

In  this  problem  the  values  of  the  reactions  are  the  same  as  those  in 
Prob.  27,  but  interchanged,  and  th  '  horizontal  component  has  the 
same  value,  but  aiitf.  in  the  opposite  direction.  See  Case  II.  of  Art.  10 ; 
therefore,  here  Hi  =  1.08.  Ri  -  2.63,  //=  1.48, 


^Jl^JiSS^-lggf 


I 


ROOF  riiUSSES. 


29 


II  all  the 

=  +  G.()8, 
=  +  2.24, 
=  -4.5, 
//'  =  cy 

),  has  its 
;'  normal 


1  normal 
he  wind 

4.6  =  he, 
.91  =  eft', 

le  dircen- 
d  load  of 
resses  in 

18  those  in 
[t  has  the 
of  Art.  10 ; 


Ans.  Stress  in  bk'  =  + 3.88,  dk'=  + 1.10,  6ft' =  -4.58 
=  eft',  h'k'  =  +  2,  ck'  =  +  2.7,  uk  =  -^  1.22,  dk  =  +  1.22,  aft 
=  -  2.91  =  eft,  hk  =  0  =  ck. 

Art.  12.  Complete  Calculation  of  a  Roof  Truss.— 

By  comparison  of  the  values  of  tlie  stresses  in  Prob.  27  uith 
those  in  Prob.  28,  we  see  that  the  stresses  are  quite  different, 
and  generally  greater  when  the  wind  blows  on  the  fixed  side 
of  the  roof  than  when  it  blows  on  the  free  side.  When  the 
wind  blows  on  the  fixed  side  it  tends  to  "  flatten  "  the  truss ; 
and  when  it  blows  on  the  free  side  it  tends  to  "  shut  up  " 
the  truss,  or  "  double  it  up." 

In  the  complete  calculation  of  a  roof  truss,  we  must  find 
the  stresses  due  to  the  greatest  dead  load,  and  combine 
them  with  the  greatest  stresses  due  to  the  live  load,  so  as  to 
get  the  greatest  possible  tension  and  compression  in  each 
member.  If  the  dead  load  and  live  load  stresses  in  any 
piece  are  of  the  same  character,  both  compressive  or  both 
tensile,  we  must  add  them  to  obtain  the  greatest  stress  in 
the  piece.  But  if  these  stresses  are  of  opposite  characters, 
one  compression  and  the  other  tension,  their  difference  will 
be  the  resulting  stress  die  to  the  combination  of  live  and 
dead  loads,  and  if  the  live  load  stress  is  less  in  amount  than 
the  dead  load  stress,  it  will  only  tend,  when  the  wind  blows, 
to  relieve  the  stress  due  to  the  dead  load  by  that  amount, 
and  the  dead  load  stress  is  the  maximum  stress  \r>  the  mem- 
ber. But  if  the  live  load  stress  is  greater  than  the  dead 
load  stress  and  of  an  opposite  character,  it  will  cause  a 
reversal  of  stress  and  this  piece  will  then  need  to  be 
counterbraced. 

It  is  the  customary  American  practice  to  determine  the 
greatest  stresses  in  each  member  of  the  fixed  side  of  the 
roof  truss  which  could  be  caused  by  the  wind  force  acting 


t 


■  t 


'  it 


•2H 


noOFS  AND  nnWGES. 


on  the  fixed  side  8  tons :  find  the  wind  stresses  in  all  the 
members 

Ans.  Stress  in  «/=  + 1 1 .16  =  ft/,  gli  =  +  8.92,  lij  =  +  C.()8, 
bg  =  -  2.5,  ch  =  -  3.10,  dj  =  -4.02,  eg  =  +1.12,  dh  =  +  2.24, 
ej  =  +  3.35,  ab  =  -9,  be  =  -7.5,  ed  =  -'6.0,  de  =  -i.5, 
kf  =:f'g'  =  g'h'  =  h'J  =  +  4.47,  b'g'  =  e'h'  =  d'j  =  0,  b'f  =  c'g' 
=  d'h'  =  0,  kb'  =  b'e'  =  c'C  =  d'e  =  -  5.0  tons. 

Prob.  27.  A  truss  like  Fig.  16,  with  one  end  free,  has  its 
span  GO  feet,  its  rise  12  feet,  the  struts  hk  and  h'k'  normal 


to  the  rafters  at  their  middle  points,  and  the  total  normal 
wind  load  on  the  fixed  sida  ac  4  tons:  find  the  wind 
stresses  in  all  the  members. 

Ans.  Stress  in  ak  =  +  5.4,  kd  =  +  2.7,  ah  =  —  4.6  =  he, 
hk  =  -2,  c&  =  +  2.7,  bk' =  + 2.7  =  dk',  bh' =  - 2.91  =  ch' , 
h'k'  =  0  =  cA:'. 

Prob.  28.  Tn  the  same  truss  Fig.  16,  with  the  same  dimen- 
sions as  given  in  Prob.  27,  let  the  same  normal  wind  load  of 
4  tons  blow  on  the  free  side  be :  find  the  wind  stresses  in 
all  the  members. 

In  this  problem  the  values  of  the  reactions  are  the  same  as  those  in 
Prob.  27,  but  interchanged,  and  th  '  horizontal  component  has  the 
same  value,  but  aiitf.  in  the  opposite  direction.  See  Case  II.  of  Art.  10 ; 
therefore,  here  Hi  =  1.08.  i?2  -  2.fl.S,  //=  1.48, 


a,BS!i»3«'.v«#^i&v«?SE*?a»*as^-;sair€  ■  ■ 


^msmmm^m^^^r 


Irv9 


ROOF  rUUSSES. 


29 


Ans.  Stress  in  bk'=+ 3.88,  dk'=  + 1.10,  6/t' =  -  4.58 
=r-.  ch',  h'k'  =  +  2,  ck'  =  +  2.7,  ctA,'  =  -^  1.22,  dk  =  +  1.22,  ah 
=  -  2.91  =  ch,  hk  =  0  =  cfc. 

Art.  12.  Complete  Calculation  of  a  Roof  Truss.— 

By  comparison  of  the  values  of  the  .stresses  in  Prob.  27  'vith 
those  in  Prob.  28,  we  see  that  the  stresses  are  quite  different, 
and  generally  greater  when  the  wind  blows  on  the  fixed  side 
of  the  roof  than  when  it  blows  on  the  free  side.  When  the 
wind  blows  on  the  fixed  side  it  tends  to  "  flatten  "  the  truss ; 
and  when  it  blows  on  the  free  side  it  tends  to  "  shut  up  " 
the  truss,  or  "  double  it  up." 

In  the  complete  calculation  of  a  roof  truss,  we  must  find 
the  stresses  due  to  the  greatest  dead  load,  and  combine 
them  with  the  greatest  stresses  due  to  the  live  load,  so  as  to 
get  the  greatest  possible  tension  and  compression  in  each 
member.  If  the  dead  load  and  live  load  stresses  in  any 
piece  are  of  the  same  character,  both  compressive  or  both 
tensile,  we  must  add  them  to  obtain  the  greatest  stress  in 
the  piece.  But  if  these  stresses  are  of  opposite  characters, 
one  compression  and  the  other  tension,  their  difference  will 
be  the  resulting  stress  due  to  the  combination  of  live  and 
dead  loads,  and  if  the  live  load  stress  is  less  in  amount  than 
the  dead  lead  stress,  it  will  only  tend,  when  the  wind  blows, 
to  relieve  the  stress  due  to  the  dead  load  by  that  amount, 
and  the  dead  load  stress  is  the  maximum  stress  in  the  mem- 
ber. But  if  the  live  load  stress  is  greater  than  the  dead 
load  stress  and  of  an  opposite  character,  it  will  cause  a 
reversal  of  stress  and  this  piece  will  then  need  to  be 
counterbraced. 

It  is  the  customary  American  practice  to  determine  the 
greatest  stresses  in  each  member  of  the  fixed  side  of  the 
roof  truss  which  could  \y)  caused  by  the  wind  force  acting 


I 


mmi 


f' 


n- 


i  i 


i 


80 


ROOFS   AND  niHbUES. 


normally  to  the  truss  on  the  fixed  side  only,  and  then  to 
build  the  members  of  the  two  sides  of  the  truss  of  the  saiuf 
size.  Since  the  stresses  caused  by  the  wind  blowing  on  the 
fixed  side  of  the  roof  are  at  least  as  great  as  those  caused 
by  the  wind  blowing  on  the  free  side,  this  arrangement 
gives  the  maximum  stresses,  and  is  on  the  safe  side;  and 
for  reasons  of  economical  manufacture  both  sides  of  the 
truss  are  constructed  alike. 

Prob.  29.  A  trust,  iikc  :  -  14,  with  one  end  free,  has  its 
span  80  feet,  its  ris'  i<;  ,<'jt,  distance  apart  of  trusses  1(3 
feet,  rafter  divided  into  four  equal  parts,  struts  vertical, 
dead  load  of  roof  20  lbs.  per  sciuare  foot  of  roof  surface,  snow 
load  20  lbs.  per  square  foot  of  horizontal  projection,  normal 
wind  load  on  fixed  side  by  table  of  Art.  '.; :  find  the  dead 
load,  snow  load,  wind  load,  and  maximum  stresses  in  all  the 
members. 

Fnmi  ^he  given  rise  and  span  we  have  the  length  of  one 
half  of  rooi  i»«  =  VIO"  +  16*  =  8V21)  =  43.08  feet. 

Weight  of  truss  =  j'^  bP  (Art.  2)  =  ^^'  ^  ^^^^)'  =  42GG  lbs. 

24 


Dead  panel  load 


420fi  ,  4.S.()8  X  1(5  X  20 


3977  lbs. 


8  4 

=  1,9885  tons,  '  .     ij,  2  tons. 
Snow  load  per  panel  =  10  x  16  ;    ."    -    200  lbs. =1.6  tons. 
Inclin.  of  roof  =  tan-'  ■  4  =  2lM<s', 
Nor.  pressure  of  wind  per  sq.  foot  (tabi  i-  A  Art.  9) = 26.2  lbs. 


Nor.  wind  load  per  panel  =  20.2  x 


43.08 


Xl6 


=  4342  lbs.  =  2.171  tons,  or  say,  2.2  tons. 

From  these  data  the  dead  lead,  snow  load,  and  wind  load 
stresses  (wind  ou  the  fixed  side  only)  may  be  computed, 


-*#3#^ee««ssteii('v>^* 


11(1  then  to 
if  tlic  .saiiif 
'ing  on  the 
ose  caused 
'i'angenieu„ 
side ;  and 
les  of  the 


ree,  has  its 
trusses  10 
is  vertical, 
[■face,  snow 
on,  normal 
I  tlie  dead 
3  in  all  the 

j;th  of  one 
t. 

42GG  lbs. 
[)77  lbs. 

=1.6  tons. 
1=25.2  lbs. 

!,  2.2  tons. 

wind  load 
computed, 


m  \ 


ROOF  TliUSSKS. 


31 


and  the  maximum  stresses  found,  for  all  the  members  of 
the  truss,  and  tabulated,  as  follows : 

StRK88K(*    in    tub    L0W£lt    ClIORI). 


Mkmiikkh. 

"/ 

fa 

(/A 

M 

Dead  lo.ad  stresses     .     . 
Snow  load  stiTSiiis     .     . 
Wind  1(  ad  stresses    .     . 

+  17.50 
+  14.00 
+  14.74 

+  17.50 
+  14.00 
+  14.74 

+  14.96 

+  11.07 
+  11.77 

+  12.44 
+  0.06 
+  8.80 

Ma:;i.uuu)  stresses     .     . 

+46.24 

+  46.24 

+  38.70 

+31.10 

Stkesses  in  the  Upper  CnoHu. 


Mkmueiui. 

ah 

be 

C'(/ 

de 

Tlead  load  stresses     .     . 
Snow  load  stresses     .     . 
Wind  load  stresses    .     . 

-18.80 
-15.04 
-12.86 

-16.10 
- 12.88 
-10,52 

- 13.40 
-10.?;; 

-   8.21 

-10.70 

-  8.56 

-  6.34 

Maximum  stresses     .    . 

-46.69 

-39.60 

-32.33 

-25.60 

Stressks  in  the  Web  Membeiis. 


ME.MIIEK8. 

«C 

dh              ej 

hg 

ch 

dj 

Dead  load  stre.sses 
Snow  load  stresses 
Wind  load  stresses 

+  1.00 
+0.80 
+  1.10 

+2.00  ]  +  6.00 
+  1.60  1  +  4.80 
+3.96  1  +  3.56 

-2.68 
-2.14 
-3.19 

-3.20 
-2.56 
-3.78 

-  3.00 

-  3.12 

-  5.04 

Maximum  stresses 

+2.99 

+  7.56  1+14.36 

-8.01 

-9.64 

-12.00 

(6/ is  not  necessary  to  the  stability  of  the  structure.) 

Here  the  maximum  stress  of  each  kind  for  each  member 
ill  the  windward  side  of  the  truss  is  found  by  adding  the 


J~lg^ 


82 


HOOFS  AM)   lilt  IDG  EN. 


(leiid  load,  snow  load,  and  wind  load  stresses  giving  the 
Rrcatost  total  tension  and  the  greatest  total  compression. 
Of  course,  the  wind  stresses  in  the  nienibors  of  the  other 
half  of  the  truss  will  be  less  than  those  above  given  for  the 
members  of  the  half  on  the  windward  side,  and  therefore, 
the  maximum  stresses  in  these  members  will  be  less  than 
those  in  the  above  table,  since  the  dead  and  snow  load 
stresses  in  the  corresponding  members  of  the  fixed  and  free 
sides  of  the  truss  are  the  same. 

If  there  is  no  wind  blowing,  the  maximum  stresses  are 
found  by  adding  together  the  dead  load  and  snow  load 
stresses.  If  there  is  neither  wind  nor  snow  the  dead  load 
stresses  are  also  the  maximum  stresses.  If  the  wind  blows 
on  the  free  side  of  the  truss,  the  maximum  stresses  cannot 
exceed  those  found  above. 

Proh.  30.  A  truss  like  Fig.  9,  with  one  end  free,  has  its 
span  90  feet,  rise  of  truss  18  feet,  rise  of  tie  rod  3  feet, 
rafter  divided  into  four  equal  parts,  struts  vertical,  dead 
load  per  panel  2  tons,  snow  load  [)er  panel  1.5  tons,  normal 
wind  load  per  panel,  wind  on  fixed  side,  2.25  tons:  find  the 
dead  load,  snow  load,  wind  load,  and  maximum  stresses  in 
all  the  members  of  the  half  of  the  truss  on  the  windward 
side.. 

Here  the  effective  reaction  for  dead  load 

=  2  X  3.5  =  7  tons. 

To  find  the  dead  load  stress  in  2-4  or  4-6,  take  moments 

around  the  point  3. 

7  X  11.25 


dead  load  stress  in  2^ : 


3.75 


:  21.00  tons. 


Similarly  the  dead  load,  snow  load,  and  wind  load  stresses 
may  be  computed  for  all  the  members  of  the  truss. 


v&v.-^-^f  ,\¥e.-.. 


:?;«■ 


cjiviiig  the 
nipression. 

the  other 
/en  for  the 

therefore, 
I  less  than 
snow  load 
(1  and  free 

tresses  are 
snow  load 
dead  load 
kind  blows 
ses  cannot 


•ee,  has  its 
od  3  feet, 
tical,  dead 
ns,  normal 
1 :  find  the 
stresses  in 
windward 


?  moments 

ons. 

id  stresses 


Ans. 


ROOF  TRUS,SES. 
Stuesses  in  the  Lower  Ciionn. 


Stresses  in  the  Uiter  Chord. 


88 


Memiiirs. 

2-» 

+  (! 

(.-•< 

S-IO 

Dead  load  8t.re.s.se8    .     . 
Snow  loiul  sti-L'sses    .     . 
Willi'  load  stresses   .     . 

+  21.00 
+  15.75 
+  18.l:j 

+  21.00 
+  16.75 
+  U,.i3 

+  18.00 
+  13.50 
+  14.40 

+  15.00 
+  11.25 
+ 10.84 

Maxiniuin  stresses    .     . 

+  .-)4.8a 

+  r,4.S8 

+  45.00 

+  37.00 

Memiikrk. 

2-3 

3-1 

5-7 

7-9 

Dead  load  stresses    .     . 
Snow  load  stresses    .     . 
Wind  load  stresses  .    . 

-22.00 
-16.05 
-16.20 

-10.30 
-14.52 
- 13.28 

-10.12 
-12.00 
- 10.20 

-12.88 

-  0.06 

-  7.74 

Mivximu  1  stress,  s    .     . 

-55.84 

-47.16 

-38.47 

-30.28 

Stresses  in 

THE    W 

'ed  Memdehs. 

Memberr. 

6-fl 

7-3 

9-10 

8-6 

6-8 

7-1  n 

Deail  load  stre.s.ses 
Snow  load  stresses 
Wind  load  stresses 

+  1.00 
+0.76 
+  1.21 

+  2.00 
+  1.50 
+  2.43 

+  7.60 
+  5.70 
+  4.54 

-3.10 
-2.32 

-3.7e 

-  3.50 

-  2.62 

-  4.23 

-  4.10 

-  3.08 

-  4.'M 

Maximum  stresses 

+2.06 

+  5.03 

+  17.84 

-9.20 

-10.35 

-12.17 

(h/is  not  necessaiy  to  the  stability  of  the  truss.) 

Prob.  31.  A  truss  like  Fig.  10,  with  one  end  free,  has  its 
span  100  feet,  its  rise  20  feet,  the  rafter  divided  into  four 
equal  parts  by  struts  drawn  normal  to  it,  dead  load  per 
panel  2.5  tons,  snow  load  per  panel  1.5  tons,  normal  wind 
load  per  panel,  wiad  on  fixed  side,  2.3  tons :  find  all  the 


ii 


II 


f 


84 


HOOFS  AND    nilllHUCS. 


stn>ssn.s  in  all  tl'.e  members  of  the  liiilf  of  the  truss  on  the 
wimlwuril  side. 


Ana. 


Sthkssus  in  the  Lowe  It  Chord. 


Mkmiikkh. 

a- 4 

4-6 

fi-IO 

Dciiil  load  stve8.sea     .     . 
Snow  load  Htressus     .     . 
Wind  load  stres-ses    .     . 

+  21.83 
+  13.10 
+  15.64 

+  18.70 
+  ll.i;2 
+  12.64 

+  12.50 
+   7.50 
+   0.21 

Maximum  stresses    .     . 

+  50.67 

+  42.40 

+  20.21 

Sthksses  in  the  UrrEit  Choisii. 


Mkmiikkh. 

2-8 

3-5 

n-T 

7-9 

Dcvd  load  stre.s.se.s    .     . 
Show  load  streasea    .     . 
Wind  load  stresses   .     . 

-23..")0 
-14.10 
-13.57 

-22.57 
—  13.55 
-13.67 

-21.66 
-12.00 
-13.67 

-20.73 
-12.44 
-13.57 

Maxiuiuin  stresses    .     . 

-61.17 

-40.09 

-48.21 

-40.74 

Stuesses  in  the  Wep  MeML'"U8. 


Mrmiiebs. 

3-4  ttiid 

l-S 

,')-•(■) 

4-fi  niKl 

0-8 

8-9 

Dead  load  stresses 
Snow  load  stresses 
Wind  load  stresses 

-2.33 
-1.40 
-2.30 

-  4.65 

-  2.79 

-  4.G0 

+  3.10 
+  1.80 
+3.11 

+  6.20 

+  3.72 
+  0.21 

+  9.30 
+  6.68 
+  0.i>l 

Maximum  stresses 

-0.03 

-12.04 

+  8.07 

+  10.13 

+24.19 

(0-10  is  not  necessary  to  the  stability  of  the  truss.) 

Prob.  32.  A  truss  like  Fig.  11,  with  one  end  fi-ee,  has  its 
spau  100  feet,  the  rise  of  truss  20  feet,  the  rise  cf  tie  rod 
2.5  feet,  the  rafter  divided  into  four  equal  parts  ly  struts 
drawn  normal  to  it,  the  dead,  snow,  and  wind  loads  2.5 


'SWRM 


tmmm 


BwiMimnnswigJWW''"'"'*'*-' 


russ  on  the 


ft-I(» 

+  12.50 
+   7.50 
+   0.21 

+  20.21 

8-9 

20 

+ 

9.30 

72 

+ 

5.58 

il 

+ 

O.iil 

13 

+24.19 

S8.) 

"ree,  has  its 

cf  tie  rod 

ts  ly  struts 

L  louds  2.5 


BOOF  TRUSSES. 


35 


tons,  1.r»  toi»s,  and  2.3  tons  respectively,  or  the  same  as  in 
I'rol).  31  :  tind  all  the  stres.'es  in  all  the  members  of  the 
half  of  the  truss  on  the  windward  side. 


Ann. 


Stkbahks  in  the  Lowkk  CiIOKI). 


.Mkmhrrh. 

L'-l 

•IB 

fi-in 

Dead  load  stre-sscs     .     . 
Snow  load  Htrt's.si's     .     . 
Wind  load  Htresses    .     . 

+  28.55 
+  17.1;{ 
+  19.92 

+  21.60 
+  11.70 
+  17.30 

+  14.35 
+  8.01 
+  7.00 

Maxiiuum  stresses     .     . 

+  05.00 

+  50.60 

+  29.96 

Stuessks  in  the  Urrmt  Chohd. 


ME.UIIKR8. 

2-8 

8-5 

6-T 

T-9 

Dead  load  stresses    .     . 
Snow  load  stresses    .     . 
Wind  load  stresses    .     . 

-30.00 
-18.12 

-19.08 
-17.81 
-18.12 

-28.75 
-17.25 
-18.12 

-27.83 
-10.70 
-18.12 

Maximum  stresses    .     . 

-67.08 

-66.61 

-64.12 

-62.65 

Stresses  in  the  Web  Members. 


Membebb, 

8-4  and 

7-8 

o-« 

4-5  and 

5-8 

6-8 

8-9 

Dead  load  stre.sses 
Snow  load  stresses 
Wind  load  stresses 

-2.33 
-1.40 
-2.30 

-  4.66 

-  2.79 

-  4.00 

+  4.00 
+  2.40 
+  4.02 

+  10.05 
+  6.39 
+  9.15 

+  14.65 

+  8.79 
+  1.".10 

Maxii.mm  stresses 

-0.03 

-12.04 

+  10.42 

+  20.19 

+30.00 

(9-10  is  not  neceasary  to  the  stability  of  the  truss.) 

Prob.  33.  A  truss  like  Fig.  11,  with  one  end  free,  has  its 
span  120  feet,  rise  of  truss  20  feet,  rise  of  tie  rod  3  feet, 
rafter  divided  into  four  equal  parts  by  struts  drawn  uornuil 


IIP" 


86 


HOOFS  AND  nUIDGES. 


to  it,  distance  lietwcon  trusses  20  feet,  dead  load  of  roof 
15  Ib.s.  per  square  foot  of  roof  surface,  siow  load  20  lbs. 
per  sijuare  foot  of  horizontal  projection,  normal  wind  load 
on  fixed  side  by  table  of  Art.  9 :  find  all  the  stresses  in  all 
the  members  in  the  half  of  the  truss  on  the  fixed  side. 

Ans.  Dead   load   per   panel  =  ,'5.1   tons;   snow  load  per 
panel  =  3  tons ;  wind  load  per  panel  =  3.3  tons. 

Stiiessks  in  the  Lower  Ciioud. 


.Mkuiicbd. 

2-4 

4-0 

8-10 

Dead  load  Htresscs     .     . 
Snow  load  stresses    .     . 
Wind  load  stresses    .    . 

+  45.26 
+  43.80 
+  37.35 

+  38.01 
+  37.65 
+  20.02 

+  22.17 
+21.46 
+  12.56 

Maximum  stresses     .    . 

+  126.41 

+  106.48 

+  66.18 

Stkerses  in  the  Uri'ER  Chord. 


MiMIIERS. 

2-:i 

S-.'i 

5-7 

T-9 

Dead  load  stresses    .     . 
Snow  load  stresses   .     . 
Wind  load  stresses   .     . 

-  47.52 

-  45.00 

-  35.31 

-  46.56 

-  46.06 

-  35.31 

-  45.60 

-  44.13 
■-  35.31 

-  44.64 

-  43.20 

-  S\3l 

Maximum  stresses    .     . 

-128.82 

-126.03 

-125.04  j  -123.16 

Stresses  in  the  Web  Memheks. 


Members. 

3-4  and 

7-S 

5-6 

4-5  and 

6-S 

6-8 

8-9 

Dead  load  stresses 
Snow  load  stresses 
Wind  load  stresses 

-2.05 
-2.86 
-3.35 

-  5.80 

-  5.70 

-  6.70 

+  6.38 
+  6.15 
+  7.44 

+  17.36 
+ 16.80 
+  17.76 

+23.72 
+  22.05 
+25.10 

Maximum  stresses 

-0.15 

-18.20 

+  10.05  j  +51.02 

+71.86 

•^'i^&^sSl^l^^^fi,^  '^V^-V^  "" 


MMMMI 


load  of  roof 
load  20  11)8. 

lal  wind  load 

itresses  in  all 

ced  side. 

ow  lo<ad  per 


T-9 

.00 
.13 
.31 

-  44.04 

-  43.20 

-  S\31 

.04 

-123.16 

J-8 

8-9 

7.36 
6.80 
7.76 

+2.3.72 
+  22.05 
+25.19 

)1.02 

+71.86 

ROOF  TltUSSES. 


87 


When  the  memhers  of  a  truss  have  many  different  in- 
clinations, as,  for  example,  in  the  Crescent  Truss  and  Arch 
Truss,  the  calculation  of  the  stresses  by  the  method  of 
moments  and  the  metluul  hy  resolution  of  forces  (Art.  (5), 
becomes  very  tedious,  owing  to  the  difficulty  in  iindinjj  the 
lever  arms,  or  the  sines  and  cosines.  In  practice  the 
graphic  method  is  often  used  for  determining  the  stresses 
in  such  trusses ;  but  if  the  analytic  method  is  i)r('ferred  to 
the  graphic,  the  truss  should  be  drawn  to  a  large  .scale,  and 
all  the  lever  arms  and  sines  and  cosines  be  scaled  from  the 
diagram. 

The  present  chapter  is  but  a  brief  treatise  on  roof  trusses. 
The  student  who  desires  to  continue  the  subject  further  is 
referred  to  more  extended  works,  such  as  "  Strains  in 
Framed  Structures,"  by  I)u  Bois,  and  "  Theory  and  Prac- 
tice of  Modern  Framed  Structures,"  by  Johnson,  Bryan, 
and  Turneaure. 


s    ; 


'I 


(IIIAPTKR    11. 

HHinr.E  THl'SSKH   WITH   UNIFOKM   LOADS. 

Art.  13.  Definitions.  —  A  Bridge  is  a  structuro  for 
carrying,  moving  loads  over  a  hotly  of  waU'r,  or  over  a 
depression  in  the  earth.  For  railroad  and  highway  acconi- 
niodations,  it  connects  two  roads  so  as  to  form  a  continnons 
road.  A  framed  brklye,  or  trussed  bridge  is  composed  of  two 
or  more  trus  ses,  which  lie  in  vertical  jjlanes,  para.lel  to  the 
line  of  the  road.  A  hridge  truss,  like  a  roof  truss,  coiisists 
of  the  upper  and  lower  chords  ai.d  the  web  members.  The 
upper  chord  of  a  simple*  bridge  tiuss  is  in  compression,  and 
the  lower  is  in  tension,  while  the  web  members  are  some  in 
compression  and  some  in  tension.  The  trusses  are  united 
by  lateral  hrachiy,  attached  to  the  panel  points  of  either  the 
upper  or  lower  ciiords,  or  both.  The  object  of  this  lateral 
bracing  is  to  sujjjjort  the  trusses  sideways,  and  stiffen  the 
structure  against  the  action  of  the  wind. 

The  Floor,  or  Floor  System,  of  a  bridge  consists  of 
tlicyfooy  beams,  stringers,  ;xrnl  floor iitrj.  'Hhejloor  beams  run 
at  right  angles  with  the  chords,  and  are  connected  to  them 
at  the  panel  points.  The  stringers  frame  into  or  rest  upon 
the  floor  beams,  and  are  parallel  to  the  chords,  and  support 
the  flooring  of  a  highway  bridge,  or  die  cross  ties  and  rails 
of  a  railroad  bridge. 

*  This  is  true  only  for  simple  trusses,  and  not  for  cantilever  trusses, 
continuous  trusses,  or  trusses  of  draw  siians. 

88 


)AI)S. 

itructuro  for 
•r,  or  over  a 
;liway  accom- 
a  cuiitiiiuous 
iposed  of  two 
ara.lel  to  the 
I'uss,  coiisists 
sinbers.  The 
prcssion,  and 
s  are  some  iu 
;3  are  united 
of  either  the 
f  this  lateral 
d  stitfen  the 

e  consists  of 
lor  beams  run 
3cted  to  them 
I  or  rest  upon 
and  support 
ties  and  rails 

ntilever  trusses, 


mtUKlK  TKUSSKS.  W 

A  Through  Bridge  is  one  in  whieli  tlio  roadway  is  sup- 
ported by  tbi'  boltdiii  clionls,  with  latenil  l)rii(!iii,^'  ovi-rhead 
bftwct'ii  tilt'  top  chords. 

A  Dock-bridge  is  one  in  whit^h  the  roadway  is  suj)- 
portt'd  by  the  toi)  chords;  tlie  tnissps  are  usnally  placed 
nearer  to  each  other  than  on  through  bridges,  the  roadway 
extending  over  them. 

When  a  tlu-otigh  bridge  is  not  of  sufficient  height  to 
allow  of  the  upper  l^nu'iil  bracing,  the  trusses  are  called 
l'())i!l  Trusses.  SiKih  trusses  are  necessarily  short ;  they  are 
"  stayed  "  by  bracing  connected  with  the  tioor  system. 

Art.  14.  Different  Forms  of  Trusses.  —  The  King, 
post  Truss  is  a  term  apiilied  to  a  truss  in  which  there  is  a 
central  vertical  tie  and  two  braces  resting  against  it,  as  in 
Fig.  17.     It  was  formerly  built  for  country  liighway  bridges 


Kits.  IT 

of  short  span.  The  term  "king-post"  is  an  old  one,  and 
came  into  use  when  the  central  piece  was  made  of  Avood, 
and  resembled  a  post,  although  its  office  was  that  of  a  tie ; 
this  tie  is  now  usually  a  wrought  iron  rod.  The  load  at  the 
center  d  is  carried  by  the  tie  up  to  the  apex  c,  and  then  by 
the  two  inclined  braces  ca  and  ch  down  to  the  abutments,  a 
and  6. 

The  Queen-post  Truss  is  a  truss  in  which  there  are 
two  vertical  ties  against  whitih  rest  two  braces,  as  in  Fig.  18. 


40 


itnoFs  A\n  iinuxsF.s. 


Tliis  is  also  an  old  term  and  has  bpcoinr  familiar  with  loiij; 
UHO.     It  is  soiiii'tiincs  cailud  u  tnipczouUil  tntH».     Tim  paiii'l 

4l 


FiB.  IH 

loads  at  h  and  />•  an'  rarricd  up  tho  tics  to  r  and  d,  and  tlu>n 
by  tliti  two  inclined  struts  m  and  til,  di.wu  to  tht!  abntnit'uta, 
a  and  h. 

The  Howe  Truss,  Fi^.  11),  has  its  voi-tic-al  niombors  in 
tension  and  the  intdincil  ones  in  coniprcssion  ;  the  dia),'onal 
counter  struts  aro  in  I     ken  lines.     The  cliouls  and  diagonal 


/. 


v/ 


Kic.  lU 

web  members  are  of  wood,  and  the  vertical  tics  of  iron. 
This  truss  was  patented  in  the  United  States  in  1840  by 
William  Howe.  It  has  proved  the  most  useful  style  of 
bridge  truss  ever  devised  for  use  in  a  new  and  tind)ered 
country.  It  is  still  very  largely  used  where  timber  is 
eheap,.for  both  highway  and  railway  bridges. 

The  Pratt  Truss,  Fig.  20,  }ms  its  vertical  members  in 
compression  and  the  inclined  ones  in  tension.     All  the 


members  of  this  truss  are  of  iron  or  steel ;  though  it  was 
formerly  built  all  of  wood,  except  the  diagonal  ties,  which 


.mVi'.mkdmmiSmm-i'iifi'M  -'■ 


1 


I  1 1 


UUIUnK  TIlUSSRS. 


41 


with  1(111!,' 
Tim  |iaii('l 


1,  iiul  then 
abntinuata, 

lonibors  in 
('  (liii>,'()iial 
III  diagonal 


B8  of  iron, 
n  1840  by 
il  stylo  of 
1  timbered 
timber   is 

lendiers  in 
,     All   the 


ngli  it  was 
;ies,  which 


worn  of  wrought  iron.  The  I'ratt  truss  is  a  favorito  tyjie, 
and  is  used  inme  tlian  any  otiicr  kin.l.  I'"i^'iire '_'(>  hIihwh  a 
f/ccA'-bridgc.  The  deck  or  ilirnugli  I'ratt  truss  is  tlie  stand- 
ard form  of  triisH  for  botli  higiiway  and  railway  bridges  of 
nio(b'rate  sjians,  though  it  is  not  generally  nse(i  for  railway 
bridges  in  whieh  the  span  is  much  less  than  lOO  feet. 

The  Warren  Truss,  or  Warren  Oirder,  iuis  all  its 
web  mend)ers  inelino;!  at  eqnal  angles,  some  of  them  being 
in  tension  and  some  in  eompression.  This  truss  is  an 
example  of  the  pine  triangular  type.  Its  web  nuMubers 
consist  always  of  ('(jiiilntenil  Iriuiiylcn.  When  the  triangles 
are  not  ecjuilateral,  the  truss  is  simply  a  ^' tridiifjiilnr  truss." 


AAAAAAAA/^ 


KiB.  ai 

Fignre  21  shows  a  Warren  truss  as  a  deck  truss.  The  War- 
ren truss  is  generally  built  all  of  iron  or  steel,  and  is  used 
for  comparatively  short  spans;  it  is  of  more  frequent 
oceurrenee  in  England  than  in  this  country. 

The  Double  Triangular,  or  Double  Warren  Truss 
(or  Oirder),  Fig.  2'J,  has  each  panel  braced  with  two 
diagonals  intersecting  each  other,  and  forming  a  single 
lattice. 


wmymmm 


Ktg.  39 


Other  forms  of  trusses  will  be  explained  as  we  proceed. 
All  these  forms  of  trusses,  and  bridge  trusses  generally, 
may  Im;  arranged  so  as  to  be  used  either  for  deck-bridges 
or  through  bridges. 


I 


^^S^Jk  . 


w 


*H 


42 


noOFS  AND  It  JUDGES. 


Art.  15.  The  Dead  Load,  or  permanent  load,  consists 
(if  tlio  entire  weij^ht  oi'  llie  hrnlge.  It  ineliitlos  the  weight 
of  the  trusses,  the  lateral  bracing,  and  the  floor  system.  In 
highivai/  biidgen,  the  floor  system  consists  of  the  floor  beams, 
which  are  supported  by  the  chords  at  the  panel  points,  the 
stringers,  which  are  supported  by  the  floor  beams,  and  the 
planks,  which  are  supported  by  the  stringers.  In  railway 
briclyes  the  stringers  support  the  cross  ties,  rails,  guard- 
rails, spikes,  etc.  The  dead  load  depends  upon  the  length 
of  the  bridge,  its  width,  its  style,  and  upon  the  live  loads  it 
is  intended  to  carry.  For  highway  bridges  with  plank 
floors  the  total  dead  load  jier  foot  may  be  found  approxi- 
mately by  the  following  formula: 

10  =  150  -f  cbl  +  4  bt, 

where  iv  =  weight  in  pounds  per  linear  foot  of  bridge. 

I  =  length  of  bridge  in  feet, 
b  =  width  of  roadway  in  feet, 
t  =  thickness  of  planking  in  inches, 
c  =  ^  for  heavy  city  bridges, 

=  ^  for  ordinary  city  or  suburban  bridges, 

=  I  for  light  country  bridges. 

For  single  track  railroad  bridges  the  dead  load  per  linear 
foot  is  given  very  closely  by  the  following  formula  (taken 
from  "Modern  Framed  Structures,"  by  Johnson,  Bryan, 
and  Turneaure,  p.  44). 

For  deck-plate  girders, 

to  =  9^4- 520 (1) 

For  lattice  girders, 

w  =  7l  +  mO (2) 


5SSS^*S^S1^'^^3|^ 


ilKlE  TRUatiES. 


43 


d,  consists 
the  weight 
'stem.  Ill 
oor  beams, 
points,  the 
s,  and  the 
[n  railway 
ils,  guard- 
tlie  knigth 
ve  loads  it 
ith  plank 
i  approxi- 


dge. 


;e8, 

per  linear 
ilae  (taken 
)n,  Bryan, 


■     •     (1) 

.     .     (2) 


For  through  pin-connected  bridges, 

v,^lH-\-  750  ...*....     (3) 

For  Howe  trusses,     iv  =  G.5 1  -\-  G75 (4) 

where  I  is  the  span  in  feet,  and  lo  the  dvjad  load  in  pounds 
per  linear  foot.  These  four  formulte  give  dead  loads  of 
iron  i-ailway  bridges,  including  an  allowance  of  400  lbs.  per 
foot  for  the  weight  of  track  material,  for  bridges  designed 
to  (iarry  lOO-ton  locomotives.  For  lighter  locomotives  the 
dead  load  would  be  less,  and  for  heavier  locomotives  it 
would  be  more.  For  double  track  bridges  add  90  per  cent 
to  the  above  values;*  for  the  load  on  each  truss  take  one 
half  the  above  values. 

Prob.  34.  What  is  the  weight  of  a  Vv'arreu  truss  bridge 
105  feet  long  ? 

Here  we  may  find  the  dead  load  from  formula  (2),  as 
follows: 

Dead  load  per  linear  foot 

=  ly  =  7  ^  +  COO  =  1336  lbs. 

.-.  weight  of  bridge  =  1335  x  105  =  140,175  lbs.  and 
weight  of  bridge  to  be  carried  by  one  truss 

=  70087.5  lbs. 

Proh.  35.   What  is  the  weight  of  a  through  pin-connected 
bridge  of  100  feet  span  ? 
Ans.  125,000  lbs. 

Proh.  36.   What  is  the  weight  of  a  Howe  truss  bridge  cf 
144  feet  span  ? 
Ans.  231,984  lbs. ;  or  115,992  lbs.  per  truss. 

♦For  double  track  brulges  the  weight  of  the  me/a/ toorA;  is  alKiutfK)% 
greater  than  for  a  single  track;  Imt  tlie  weight  of  tlie  track  materiat  is 
just  double  that  for  a  single  track  bridge. 


i! 


ill     i 


• -"-""rKBaSe*. 


44 


HOOFS  AND  nillDGES. 


Art.  16.  The  Live  Load,  or  moving  load,  is  that  which 
moves  over  thft  bridjjo,  ami  consists  of  \va<?ons  ami  foot  pas- 
sengers on  highway  luidg-s,  and  trains  on  railway  bridges. 

For  highwatj  brirhjes,  the  live  load  is  usnally  taken  as  a 
uniform  load  of  from  r)()  to  100  lbs.  per  square  foot  of  road- 
way, or  the  heaviest  concentrated  load,  due  to  a  road-roller, 
which  is  likely  to  come  upon  the  bridge.  The  trusses  of 
highway  bridges  are  usually  found  to  receive  the  greatest 
stresses  from  a  densely  packed  crowd  of  people,  while  the 
Hoor  system  usnally  rec^eives  the  greatest  stresses  from  the 
concentrated  load.  The  following  are  the  live  loads  speci- 
fied by  Waddell,  per  sipuire  foot  of  floor: 


Span  of  IliltlBo. 

C'tty  and  Suburban  nrlitgcs. 

Ciiuntry  llridKf!). 

0  to    50  feet. 

100  lbs. 

OOlba. 

60  to  150  feet. 

00  lbs. 

80  lbs. 

150  to  200  feet. 

80  lb" 

70  lbs. 

200  to  300  feet. 

70  lbs. 

60  lbs. 

300  to  400  feet. 

60  lbs. 

60  lbs. 

The  live  load  per  panel  of  a  highway  bridge  may  then  be 
found  by  multiplying  the  width  of  roadway  including  the 
sidewalks  by  the  product  of  the  load  per  square  foot  with 
the  panel  length. 

For  railway  bridgen  the  live  load  is  o^  en  specified  as  the 
weight  of  two  of  the  heaviest  locomotives  which  we  think 
will  ever  pass  over  the  bridge,  and  followed  by  a  uniform 
load  due  to  the  heaviest  possible  train. 

In  English,  and  sometimes  in  American  practice,  it  is 
considered  sufficiently  accurate  to  use  the  corresponding 
uniformh'  (ilmributed  load  which  causes  the  same  stresses 
in  the  chords  as  those  which  result  from  the  above  con- 
centrated locomotive  loads. 


it  which 
'oot  pilH- 
)riilg<'s. 
:en  as  a 
of  road. 
,d-roller, 
usses  of 
greatest 
hile  the 
rom  the 
Is  speci- 


iiltfes. 

i. 
i. 

i. 


then  be 
ling  the 
oot  with 

i  as  the 
i^e  think 
uniform 

ce,  it  is 

jponding 

stresses 

ove  con- 


nitlDGE  TRUSSES. 


4r) 


Tlio  following  etjuivalent  uniformly  distri'i>iited  live  loads 
per  linear  foot  of  track,  i)roduce  approximately  the  same 
stresses  as  the  above  concentrated  loads. 

Span,  10,  20,  30,  40,  50,  100,  200,  300  feet. 

Load,  10,000,  6600,  5500,  4900,  4600,  4000, 3700, 3500  lbs. 

This  loading  is  very  nearly  that  used  by  the  rcnnsyl- 
vania  Railroad.  The  Erie  Railroad  nses  a  live  load  one  fifth 
greater;  and  the  Lehigh  Valley  Railroad  a  live  load  one 
third  greater. 

The  live  load  is  taken  greater  for  short  spans  than  for 
long  ones,  because  if  the  span  is  short  one  t)r  two  locomo- 
tives may  cover  the  whole  bridge,  while  if  the  span  is  long, 
the  whole  bridge  would  not  often  be  loaded  with  more  than 
a  train  drawn  by  one  or  two  locomotives.  'J'he  calculations 
of  the  stresses  are  made  in  precisely  the  same  way  for 
railway  as  for  highway  bridges,  so  long  as  the  live  load  is 
uniform. 

Ill  the  use  of  equivalent  uniform  loads,  the  following  precautions 
are  to  be  observed :  In  obtaining  the  stresses  in  the  chords  hrA  the 
main  web  members,  the  equivalent  load  corresponding  to  the  length 
of  the  truss  is  to  be  used.  But  in  such  members  as  receive  a  maxi- 
mum stress  from  a  single  panel  load,  another  equivalent  load  must  be 
used.  Thus,  in  the  "hip  verticals"  of  a  Pratt  truss,  Fig.  30,  or  in 
the  "vertical  suspenders"  of  a  Warren  truss.  Fig.  35,  the  maximum 
stress  is  obtained  by  using  the  equivalent  load  corresponding  to  a 
span  of  two  panel  lengths. 

Prob.  37.  A  bridge  for  a  city  has  its  roadway  20  feet 
wide  in  the  clear,  and  also  two  sidewalks,  each  6  feet 
wide  in  the  clear.  The  span  is  200  feet  and  there  are  10 
panels.     Find  the  live  panel  load  per  truss. 

Ans.  12.8  tons. 


HI 

i  H 


I    i  I 


V'i 


mm 


46 


ROOFfI  AND   niilDGKS. 


Fig.  S3 


O, 


Pinb.  38.  A  country  l):i(lj,'C,  .50  feet  long,  lias  its  roadway 
1(5  feet  wide  in  the  dear,  and  a  sidewalk  8  feet  wide  in  tlie 
clear.  There  are  5  panels.  Find  the  live  panel  load  per 
truss. 

Ans.  6.4  tons. 

Art.  17.  Shear — Shearing  Stress.  —  Let  Fig.  23  repre- 
sent a  beam  fixed  horizontally  at  one  end  and  sustaining  a 

load  P  at  the  other  end.  Imagine 
the  beam  divided  into  vertical  slices 
or  transverse  sections  of  small  thick- 
ness. The  weight  /*  tends  to  sepa- 
rate or  shear  the  section  or  slice  on 
which  it  immediately  rests  from  the 
adjoining  one.  The  lateral  con- 
nection of  the  sections  prevents  this 
separation,  and  the  second  section  or  Siice  is  drawn  by  a 
vertical  force  ecpial  to  the  weight  P  which  tends  to  slide  or 
shear  it  from  the  third  section,  and  so  on.  Thus,  a  vertical 
force  equal  to  the  weight  P  is  transmitted  from  section  to 
section  throughout  the  length  of  the  beam  to  the  point  of 
support.  This  vertical  force  is  called  the  "  shearing  force," 
or  ^' shear" ;  and  the  equal  and  opposite  internal  force  or 
stress  in  the  section  that  balances  it  is  called  the  "  shearing 
stress." 

The  shear  then  at  any  section  is  that  force  ivhich  tends  to 
make  that  section  slide  upon  the  one  imtnediatel;/  following. 

The  vertical  shear  at  any  section  is  the  total  vertical  force 
at  that  section,  and  it  is  equal  to  the  sum  of  the  vertical  com- 
ponents  of  all  tl  external  forces  acting  upon  the  beam  on 
either  side  of  the  section. 

Thus,  let  Fig.  24  represent  a  beam  I  feet  long,  resting 
horizontallj'  on  supports  at  its  extremities;   and  let   the 


'CS:r7SrvS(Sr^T^-fi 


%i 


BIIIDGE  TRUSSEi.. 


47 


beam  have  a  uiiifonn  l(ni(l  of  to  lbs.  i)or  foot.  Consider  a 
section  ab  at  any  distance  a;  from  the  left  end.  Then,  the 
reaction  at  each  siii)i)ort  is  J  wl;  the  slicar  in  the  section 
ab  is  the  weiglit  that  is  between  that  section  and  tiic  center 


Hiw. 


yilvi. 


Fig.  S4 


of  the  beam,  or  (|  I  —  x)w,  since  this  is  the  total  vertical 
force  at  ab,  wliich  is  the  shear  bi'  definition;  i)ut  tliis  is  the 
same  as  the  algebraic  sum  of  the  vertical  components  on 
the  left  side  of  ab. 

For  the  algebraic  sum  of  the  vertical  components  on  the 
left  of  the  section  ab  =  ^  Iw  —  xw,  or  (^  I  —  x)w. 

Art.  18.  Web  Stresses  due  to  Dead  Loads  —  Hori- 
zontal Chords. —  Let  ABCD  be  a  truss  with  horizontal 

c     cT  /»     fc     h'  e'   D 


-4    «     b  y,  « 


d     c '    jt>'   a 


7-^B 


P. 

Fig.  SB 

chords*  AB,  CZ),  and  let  a  section  mn  be  drawn  in  any 
panel  be  cutting  the  diagonal  bh  and  the  two  chords,  and 
let  6  be  the  angle  which  the  diagonal  nuvkes  with  the  ver- 
tical. Then  (Art.  5)  the  stresses  in  these  three  members 
mist  be  in  equilibrium  with  the  external  forces  on  the  left 
of  the  section;  and  therefore   the  algebraic  sum  of  the 

.  *  Called  also  Ihui^bs. 


K 


T? 


48 


nooFs  AND  liiiinaKs. 


w 

m 


vertical  components  of  theso  forces  (Mjiials  zero.     Hence, 
calling  S  the  stress  in  the  diagonal  bh,  wt;  have 

R-Pi-I\  +  Scos6  =  0 (1) 

But  It— Pi  —  Pj,  is  the  vertical  shear  for  the  given  socticjn 
Miw  (Art.  17). 
Hence  (1)  becomes 

Shear +  5008(9=0; 

.".  S  =  —  shciiT  sec  0 (2) 

Therefore,  for  horizontal  chords  and*  vertical  loads,  the 
stress  in  any  web  member  in  etjmd  to  the  vertical  shear  multi- 
plied by  the  secant  of  the  angle  which  the  member  makes  tvith 
the  veHical. 

Since  the  shear  in  the  section  cutting  he  is  the  same  as 
that  in  the  section  mn,  no  load  being  at  h,  therefore  the 
stress  in  Ch  is  equal  to  the  shear  in  ch,  that  is,  equal  to  the 
shear  in  the  panel  be.  Since  there  are  no  loads  between 
the  joints  h  and  c,  the  shear  is  constant  lluoughout  the 
panel  be;  and  we  nsually  speak  of  it  as  the  shear  in  the 
panel  be. 

It  will  be  observed  from  (2)  that,  for  the  diagonal  bh, 
the  stress  is  negative,  or  compressive,  provided  that  the 
shear  is  positive;  but  for  the  dead  load  the  shear  is  always 
positive  in  sections  left  of  che  middle  of  the  truss.  Hence 
the  stress  in  b'  is  compressive,  and  so  for  all  the  other 
diagonals  in  th .  left  half  of  the  truss,  since  they  are  all 
inclined  in  the  same  direction,  that  is,  downwards  toward 
the  left.  Conversely,  for  members  inclining  downwards 
toward  ,  he  right,  in  the  left  half  of  the  truss,  the  stresses 
axe  2>ositive  or  tensile. 

Prob.  39.  A  throngh  Howe  truss,  like  Fig.  2-'i,  has  8 
panels,  each  15  feet  long,  aud  20  feet  deep:  find  all  the 


i 


i 


BltlDGE  TliUSSES. 


49 


web  stresses  due  to  a  dead  load  of  450  lbs.  per  linear  foot 
per  truss. 

Panel  load  =  1^0^15  =  27  ^  3  gg  ^^^^ 
2000  8 


27     7 
Reaction    =--  y.^ 


=  ^  =  11.8  tons, 
16 


which  is  also  the  shear  in  panel  Aa.  The  shear  in  each  of 
the  other  panels  is  found  by  subtracting  from  the  reaction 
the  loads  on  the  left  of  the  panel. 

sec<?=-    — ^-=j  =  1.26. 

The  secant  for  the  verticals  =  1. 

We  have  then  the  following  stresses  for  the  diagonals : 

Stress  in  AC^-  11.8  x    1.25  =  -  14.8  tons. 

Stress  in  oe  =  -  (11.8  -  3.38)1.25  =  -  10.5  tons. 
Stress  in  67i  =  -(11.8-  6.76)1.25  =  -  6.3  tons. 
Stress  in    cfc  =  -  (11.8  -  10.14)1.25  =  -    2.1  tons. 

For  the  verticals: 

Stress  in  aC=  + 11.8  tons.     Stress  in  be.  =  +  8.4  tons. 
Stress  in  ch  —  +   5.1  tons.     Stress  in  dk  =  +  3.4  tons. 

This  last  value  is  found  by  passing  a  section  around  d  cut- 
ting dc,  dk,  dc'  (see  Rem.  of  Art.  5),  and  taking  vertical 
components.  #  ' 

Prob.  40.   A  deck  Pratt  truss,  like  Fig.  20,  has  12  panels, 
each  6  feet  long,  and  6  feet  deep:  find  all  the  web  stresses 
due  to  a  dead  load  of  500  lbs.  per  linear  foot  per  truss. 
Ans.   Stresses  in  verticals 

=  _8.25,  -6.75,  -5.25,  -3.75,  -2.25,  -1.50  tons. 
Stresses  in  diagonals 

=  +  11.63,  +9.61,  +7.40,  +5.28,  +3.17,  +1.05  tons. 


■•^""siaa. 


m 


60 


nooKS   AM>    lilillXiES. 


/'loh.  41.    A  throujrh  Warroii  truss,  Fig.  liCi,  lias  10  piincls, 
(uu'li  10  feet  long,  its  web  uiuinbeis  all  loruiiiig  I'tiuilatoral 

t         s         r,         7         [t         ;i'        7'       r'        s'        f 

/\/.\/.\/\/\/\AA/\A 


10  It         10' 

Ki«,  SO 


4'         t' 


triangles  (Art.  14):  find  the  stresses  in  all  the  web  mcnihers 
dno  to  a  cleatl  load  of  400  lbs.  per  linear  foot  per  truss. 

Ans.  Stresses  in  1-2,  ,">-4,  5-0,  7-8,  9-10  are  -  10..'{«, 
-  8.08,  ~r>.7(i,  -  3.4(5,  -  1.14  tons. 

Stresses  in  1-4,  3-0,  5-8,  7-10,  9-12,  are  +  10.38,  +  8.08, 
+  .5.70,  +  3.40,  +  1.14  tons ;  that  is,  the  signs  alternate  in 
the  web  members. 

Art.  19.  Chord  Strasses  due  to  D«>.ad  Loads  — 
Horizontal  Flanges. 

(1)  To  find  the  chord  stresses  by  the  method  of  moments. 

Pass  a  section  cutting  the  chord  member  whose  stress  is 
required,  a  web  member,  and  the  other  chord,  and  take  the 
center  of  moments  at  the  intersection  of  the  web  member 
and  the  other  chord.  Then,  supposing  the  right  part  of  the 
truss  removed,  state  the  equation  of  moments  between  the 
unknown  stress  and  the  exterior  forces  on  the  left  of 
the  section.  For  stresses  in  the  npper  chord  members  the 
centers  of  moments  are  at  the  lower  panel  points;  and  for 
stresses  in  the  lower  chord  members  the  centers  are  at  the 
upper  chord  points. 

Thus,  in  Vrob.  41,  each  panel  load  is  2  tons,  and  each 
reaction  is  therefore  9  tons;  the  depth  of  the  truss  is 
10  sin  00°  =  8.66  feet.  Hence  for  the  lower  chord  stresses, 
we  have : 


-«v 


IS  10  piint'l.s, 
i  i'(liiilatoral 

s' 1' 

4'         f 


kX'b  inciiihers 
jr  truss. 

we  -  1().,'}«, 

).38,  +  8.08, 
alternate  in 


1  Loads  — 

moments. 

086  stress  is 
and  take  tlie 
veb  member 
t  part  of  tlie 
between  the 
tlie  left  of 
nembers  the 
nts;  and  for 
I's  are  at  the 

IS,  and  each 
the  truss  is 
ord  stresses, 


BRIDGE  riiUSSES.  61 

9  X  5  -  stress  in  '2-4  x  8.(i(;  —  0; 

.-.  Stress  in  2-4  =  +  r».li()  tons. 
9x15-2x5-  stress  in  4-0  x  H.m  =  0; 

.-.  Stress  in  4-G  =  +  14.4.'!  tons. 
9  X  25  -  2(15  +  5)  -  stress  in  O-S  x  H.GG  =  0; 

.-.  Stress  in  0-8  =  +  21. .'50  tons. 
9  X  35  -  2  X  45  -  stress  i!i  8-10  X  8.00  =  0; 

.-.  Stress  in  8-10  .=  +  25.98  tons. 
9  X  45  -  2  X  SO  -  stress  in  10-12  x  8.00  =  0; 

.■•  Stress  in  10-12  =  +  28.28  tons ; 

and  for  the  upper  chord  stresses,  we  have : 
9  X  10  -  stress  in  1-3  x  8.00  =  0 ; 

.-.  Stress  in  1-3  =  -  10.38  tons. 
9  X  20  -  2  X  10  -  stress  in  3-5  x  8.06  =  0 ; 

.-.  Stress  in  3-5  =  - 18.48  tons. 
9  X  30  -  2(20  +  10)  -  stress  in  5-7  x  8.06  =  0; 
.  .-.  Stress  in  5-7  =  —  24.24  tons. 

9  X  40  -  2  X  00  -  stress  in  7-9  x  8.06  =  0; 

.-.  Stress  in  7-9  =  —  27.70  tons. 
9  X  60  -  2  X  100  -  stress  in  9-9  X  8.60  =  0; 

.-.  Stress  in  9-9  =  - 28.86  tons. 

(2)  By  the  method  of  chord  increments. 

Let  it  be  required  to  find  the  stress  in  the  chord  member 
0-8,  ¥\^.  20.  Denc^te  the  vertical  shears  in  the  web  mem- 
bers 2-1,  1-4,  4-^},  etc.,  by  ?'i,  r^,  v^,  etc.,  and  the  angles 
these  members  make  with  the  vertical  by  fl„  0^,  6^,  etc. 
Now  pass  a  curved  section  cutting  the  chord  member  6-8 
and  all  the  web  members  on  the  left.     Then  from  the  first 


i 


ar" 


-^raai»^»r/8»«teTf?^V'«E!s?jg5p«fvs:aE^^ 


62 


ROOFS  AND   HlUbGEs. 


coiulition  of  «Mi\iilibri»iiii  the  sum  of  the  hoiizontal  rom- 
poncnts  is  zero.  Hut  the  horizontal  eoinpoiiont  of  tlio 
8tresH  in  any  wob  nicnilu'r  is  0(iual  to  tlio  vertical  shear  in 
that  nioniher  iniiltii)lied  by  the  tangent  of  its  angle  with 
the  vertical.  Hence  we  have: 
Stress  in  G-S 

=  , ,  tan  0,  +  ih  tan  61 4- 1\  tan  6^  +  ('« tan  ^4  +  I'j  tan  d.,. 
'riicirjhn;  to  Jind  the  stn-itu  in  uny  chord  member,  imiss  a 
riirri'il  ncHiou  ckUiiiij  the  member  ami  all  the  web  membeh  on 
the  left,  mnllliihl  the  uertinU  shear  in  vin'h  ireh  member  by  the 
taiiijeid  of  its  amjle  with  the  vertical,  and  take  the  sum  of  the 
jtroihtcts. 

From  an  inspection  of  Fig.  26  it  is  seen  that  the  stress  in 
any  chord  member,  as  4-6  for  example,  is  greater  than  the 
stress  in  the  immediately  preceding  chord  member,  2-4,  by 
the  sum  of  the  horizontal  components  of  the  stresses  in  the 
web  members,  1-4  and  4-3,  intersecting  at  the  panel  point 
between  these  two  members ;  That  is,  the  increment  of  chord 
stress  at  any  panel  point  is  equal  to  the  sum  of  the  hqrizontal 
com2>onents  of  the  web  stresses  intersecting  at  that  jmint ;  or, 
expial  to  the  sum  of  the  jn-oducts  of  the  vertical  shears  of  these 
web  members  by  the  tangents  of  their  respective  arujles  ivith  the 

vertical. 

It  is  well  to  test  the  stress  found  by  this  method  by  the 
method  of  n.oments,  as  the  results  obtained  by  the  two 
methods  should  agree ;  and  this  affords  a  check  on  the  work. 

The  method  by  chord  incremants  applies  only  to  hori- 
zontal chords,  while  the  method  by  moments  applies  to 
trusses  of  any  form,  that  is,  when  the  chords  are  not  hori- 
zontal as  well  as  when  they  are. 

Prob.  42.  Let  it  be  required  to  find  the  chord  stresses  in 
Prob.  41  by  the  method  of  chord  increments. 


•izontal  <'<>iu- 
lUPMt.  of  tlio 
iciil  Hhciir  in 
8  angle  with 


+  1-4  tan  6y 

ember,  }Htns  a 
h  members  on 
:„ember  b;/  the. 
he  sum  of  the 

t  the  stress  in 
[vter  than  the 
niber,  2-1,  by 
itresses  in  the 
e  panel  point 
ime)U  nf  chord 
the  horizontai 
hat  2)oint ;  or, 
shears  of  these 
angles  ivith  the 

nethod  by  the 
tl  by  the  two 
k  on  the  work. 
,  only  to  hori- 
its  applies  to 
!  are  not  hori- 

3rd  stresses  in 


i\ 


t 


lUilhOK  TUUSSEfi. 


58 


Hero  each  panel  load  is  2  tons,  and  all  the  web  members 
are  inclined  at  an  aiiK'le  of  ',\0°; 

.-.  tan  Ox  =  tan  0^  =  etc.  =  tan  IW  =  .5773. 
Hence  for  the  lower  chord  stresses  we  have: 
Stress  in  2^  =  9  x  .r.773  =  +  r>.2(). 
Stress  in  4-<5  =(9  +  1)  +  7)  x  ..".773  =  +  14.43. 
Stress  in  G-8  =(9+9  +  7+7  +  5) x.fi773=  +21.36. 
Stress  in  8-10  =(32+5+5+3) X. 5773=  +25.98. 
Stre,ss  in  10-12  =(42+3 +3 +  l)x. 5773= +28.29; 

and  for  the  upper  chord  stresses  we  have: 

Stress  in  1-3  =  -(9  +  9)  X  .5773  =  -  10.39. 

Stress  in  li-o  =  -(18+2  x  7)  X  .5773=  -18.47. 

Stress  in  5-7  =  -(32  +  2  x  5)  x  .5773=  -24.25. 

Stress  in  7-9  =  -(42  +  2  x3)  x .5773  =  -27.71. 

Stress  in  9-9  =  -  (48+2  x  1 )  x  .5773  =  -  28.80. 

I^ob.  43.   Find  all  the  chord  stresses  in  the  deck  Tratt 
truss  of  I'rob.  40. 
Ana.   Stresses  in  lower  chords 

=  +  8.25,  +15.0,  +20.25,  +24.0,  +20.25  tons. 
Stresses  in  upper  chords 

=  -8.25,  -15,  -20.75,  -24,  -20.25,  -27.0  tons. 

Prob.  44.    A  through  Howe  truss,  like  Fig.  25,  has  12 

panels,  each  10  feet  long,  and  10  feet  deep :  find  the  stresses 

in  r.ll  the  members  due  to  a  dead  loa^l  of  400  lbs.  per  linear 

foot  per  truss. 

Ans.    Lower  chords=11.0,  20.0,  27.0,  32.0,  35.0, 30.0  tons. 

Verticals    =11.0,  9.0,  7.0,  5.0,  3.0,  2.0  tons. 

Diagonals  =15.5,  12.08,  9.80,  7.04,  4.22,  1.4  tons. 

Of  course  the  upper  chord  stresses  can  be  writtpu  directly 
from  those  of  the  lower  chord  by  (1)  of  Art.  5. 


Wi»»si?!pct.;Kj»7aR(?<SK^<'!i=s»-3T*.»'Wjai»rm5tjiss«wn.<x-jra^^ 


54 


ROOFH  AS1>   HUWaES. 


Art.  20.    Position  of  Uniform  Live  Load  Causing 
Maximum  Chord  Stresses.—  Let  Kig.  U7  bo  a  truss  sup- 


I      .1 


s 


1\ 


1       n      tt 


N 


FiB.  07 


ii        Ik 
III. 


ported  at  the  oiuls.  Tl.en  t(.  fiiitl  the  stress  in  any  chord 
ineuiber,  we  i>iiss  a  section  cutting  that  member,  a  web 
member,  and  the  other  chord,  and  take  the  center  of 
moments  at  (he  iniersectiim  of  the  web  mcmbvr  and  the 
other  chord;  and  since  tiie  hn-er  arm  for  the  chord  is  con- 
stant,  tlie  stress  in  any  chord  mend)er  will  be  greatest  when 
the  live  load  is  so  arranged  as  to  give  the  greatest  bending 

moment. 

Now  siippose  we  have  a  uniformly  distributed  moving 
load  coming  on  the  truss  from  the  right,  till  it  produces  the 
left  abutment  reaction  Ru  then  any  increase  in  the  load  on 
the  right  of  the  section  .V  will  artVct  the  forces  on  the  left 
only  by  increasing  the  reaction  U„  and  consequently  the 
bending  moment.  Hence,  as  R,  increases  with  every  load 
added  to  the  right  of  .V,  the  bending  moment  increases,  aiul 
therefore  the  chord  stress  also  increases.  Also,  suppose  we 
have  a  uniformly  distributed  moving  load  covering  the  truss 
on  the  left  of  the  section  producing  the  right  abutment 
reaction  R.t\  then  any  increase  in  the  load  on  the  left  of  the 
section  N  will  affect  the  forces  on  the  right  only  by  increas- 
ing the  reaction  It,,  and  consequently  the  bending  moment. 
Hence  every  load,  whether  on  the  right  or  left  of  the  sec- 
tion, increases  ths  bending  moment,  and  therefore  the  chord 
stress. 


t-!^^- 


id  Causing 

!  ;i  truss  Slip- 


in  any  chord 
uber,  11   wcl) 

10  centt'i'  of 
i\\hv  aail  the 
flioid  is  con- 
jreatt'st  wlicii 
itest,  bendiii;,' 

lutnd  moving 
prodiicos  tho 

11  the  h)ad  on 
9  on  tho  left 
jeqnently  the 
th  every  h)ad 
increases,  and 
0,  suppose  we 
ring  the  truss 
;ht  abutment 
:he  left  of  the 
ly  by  increas- 
ing moment, 
ft  of  the  sec- 
Fore  the  chord 


I 


hlillXlK   TliVSHFU. 


66 


Thnrfnre,  for  a  uniform  had,  thv  maximum  hpmli»o 
,„nmnd  at  'au'i  paiut,  a>ul  cum'r'uHn  thv  maximum  rlnml 

Hlrr^x  in  «»// ' mrmhvr,  » irx  whru   Ihv  live  Imd  corers  the 

whole  lenyth  of  the  truss. 

To  determine  tho  chord  stresses  then  due  to  a  uniform 
live  load,  we  have  only  to  sui.pose  th.*  live  load  to  cover 
tlu*  whole  truss,  just  as  the  .lead  h.a.l  does,  au.l  compute  tho 
cliord  stresses  in  exactly  the  sauie  way  as  wo  compute  tho 
dead  load  stresses. 

Prob.  46.  A  through  Warren  truss,  like  Fig.  2C>,  has  8 
panels,  each  8  feet  long:  find  the  stresses  in  all  tho  chord 
mend)ers  duo  to  a  live  load  .)f  1000  lbs.  per  linear  foot  per 

truss. 

Ans   Upper  chord  stresses=ir,.lG,  27.72, 34.04,  30.0(>  tons. 
Lower  chord  strcsses=  8.08,  21.92,  31.20,  35.80  tons. 

Prob  46.  A  deck  Tratt  truss  has  11  panels,  each  11  feet 
long,  atul  11  feet  deep:  find  all  tho  chord  stresses  due  to  a 
live  load  of  800  lbs.  per  linear  foot  per  truss. 

Ans.   Upper  chord  stresses 

=  22.0,  39.6,  52.8,  61.6,  66.0,  66.0  tons. 

Art  21.  Maximum  SttesseB  in  the  Chords.— Accord- 
in"  to  "the  principles  of  tho  preceding  Article  the  stresses  in 
the  chords  will  be  greatest  when  both  dead  and  live  loads 
cover  tho  whole  truss.  We  have  then  only  to  determine 
the  stress  in  each  chord  merabDr  due  to  the  dead  and  live 
loads,  as  in  Arts.  19  and  20,  and  take  their  sum;  or,  we  may 
add  together  at  first  the  dead  and  live  panel  loads,  and 
determine  tho  maximum  stress  in  each  chord  member 
directly.  This  method  is  the  simplest  and  shortest;  but 
it  is  not  the  one  which  in  practice  is  generally  employed. 


•'^mmssssmET 


--..-.■vj^'mm*. 


fi.-5SGS1^P" 


66 


nOOFS  AND  BRIDGES. 


.■.-.  J, 


=  2J  sq.  in. 


All  modern  specifications  require  a  separation  of  dead  and  live 
load  stresses.  One  type  of  specilication  pernjits  twice  the  allowable 
stress  per  s(puivo  incli  of  niital  for  dead  load  that  it  permits  for  live 
load;  tluis  1(),0()0  lbs.  per  S(iuare  inch  lor  live  load  stresses  and 
20,000  lbs.  per  square  inch  foi  (lea<l  load  stresses.  So  that  if  a  tension 
member  ha.s  a  live  load  stress  of  100,000  lbs.  and  a  dead  load  stress  of 
60,000  lbs.,  the  sectional  area  is  determined  as  follows  : 

ioooo_o^io       „,      ^. 

10000  20000 

That  is,  12i  square  inches  are  required  for  the  sectional  area  of  the 
member. 

Another  type  of  specification  re(iuires  the  sectional  area  to  be 
determined  by  a  consideration  of  the  minimum  and  maximum  stressea. 
Thus,  for  tension  members,  the  allowable  stress  is 

10000  fl  +  niHL_fi!!^?\lbs.  per  square  inch. 
V        max.  stress  / 

In  the  above  case  the  allowable  stress  would  be 

10000 /l  +  -5^  ^  =  13333  lbs.  per  square  inch. 
V        16(K),Hi; 

Proh.  47.  A  (lock  Howe  truss,  for  a  railroad  bridge,  Fig. 
28,  has  12  panels,  each  12  feet  long,  and  12  feet  deep;  the 


9     11 


li   IV  9'   T    r<'    a'    I' 


~t    6     8    io    la   ih   u'  io'  a'   O'    h' 

Kig.  S8 

dead  load  is  given  by  formula  (4),  Art.  15,  the  live  load  is 
laOO  lbs,  per  linear  foot  per  truss :  find  the  maximum  chord 
stresses. 

Here  we  have  the  dead  panel  load  per  truss  from  for- 
mula (4)  , 

_ (6^J444i675)il2 ^ g^j^g  y^^  _^  ^ 333  ^^^g^ ^^^  5  t^^g 

Live  panel  load  per  truss  =^|j^^=9  tons. 


?a(l  and  live 
he  allowablo 
mils  for  live 
stresses  and 
t  if  a  tension 
load  stress  of 


il  area  of  the 

1  area  to  be 
mum  stresses. 


bridge,  Fig. 
t  deep;  the 
1' 


1, 


live  load  is 
imum  chord 

83  from  for- 

3,  say  6  tons. 
I  tons. 


i 


^ 


,-««fi»  ^ 


f? 


BlilDGE  TRUSSES. 


67 


We  have  then  at  each  upper  apex  a  load  of  5+9=14  tons, 
from  which  tho  maximum  cliord  stresses  may  be  coj'-i  "ted 
directly. 

Ana.   Maximum  stresses  in  lower  chord 

=  77, 140,  189,  224,  245,  252  tons. 
The  upper  chord  stresses  can  be  written  from  the  lower 
by  (1)  of  Art.  5. 

Prob.  48.   A  deck  Howe  truss  has  11  panels,  each  il  feet 
Ion",  and  11  feet  deep;  the  dead  load  is  400  lbs.  per  foot 
per°trus3,  and  the  live  load  is  1200  lbs.  per  foot  per  truss: 
find  the  maximum  chord  stresses. 
Ans.   Lower  chord  stresses 

=  44.0,  79.2,  105.6, 123.2,  132,  132  tons. 
The  upper  chord  stresses  for  the  Howe  and  Pratt  trusses 
equal  the  lower  chord  stresses  with  contrary  signs,  (1)  of 
Art.  5;  also,  all  the  chord  stresses  in  the  Howe  and  Pratt 
are  the  same,  no  matter  on  which  chord  the  loads  may  be 
placed. 

Art.  22.  Position  of  Uniform  Live  Load  Causing 
Maximum  Shears.  — (1)  For  a  uniform  live  load  the 
maximum  positive  shear  at  any  point  N,  Fig.  27,  occurs 
when  every  possible  load  is  added  to  the  right  and  when 
there  is  no  load  on  the  left;  for  the  shear  at  this  point  N  is 
then  equal  to  the  left  reaction  B^,  and  adding  a  load  to  the 
right  increases  the  left  reaction  E^  and  therefore  the  posi- 
tive shear,  while  adding  a  load  to  the  left  decreases  the 
positive  shear.  Hence  the  viaxiumm  positive  shear  at  any 
point  occurs  ivhen  the  live  load  extends  from  that  point  to  the 
remote  abutment. 

(2)  Let  a  uniform  live  load  be  placed  on  the  left  of  N, 
producing  the  left  reaction  R^.     Now  this  reaction  R^  is 


I 


"imi' 


*w*    -'"-*^'**«era«l!*K*ft, 


}Si^^^^3m^-x-iSfs^si^-4^^miiiSiSif!m-'mvs^ss^.T^^^i&^f^ff^s^i(t;!^i^m^ir^' 


'.f 


68 


ROOFS  AND   nniDGES. 


only  ii  part  of  the  weight  of  the  live  load,  while  the  shear 
is  this  roai'tion  niimis  the  whole  weight  of  the  load,  and  is 
therefore  )ii'(jiUire ;  and  every  load  added  to  the  left  of  this 
point  increases  nunierieally  this  negative  shear.  Hence  the 
maximum  tmjatioe  shear  at  any  point  occurs  ivhen  the  live  load 
extends  from  that  point  to  the  nearest  abutment. 

Prob.  49.  Find  the  greatest  positive  and  negative  shears 
for  each  panel  of  Fig.  27  when  the  apex  live  load  is  9  tons. 

hy  the  principle  just  proved,  the  greatest  positive  live 
load  shear  in  any  panel  occurs  when  all  joints  on  the  right 
are  loaded,  the  joints  on  the  left  being  unloaded.  Hence, 
for  greatest  positive  shear  in  1-3  the  truss  is  fully  loaded. 
We  have  then 

Positive  shear  in  1-3  =  9x2^  =  22.5  tons. 

There  is  no  negative  shear  in  1-3. 

For  greatest  shear  in  3-5  all  joints  except  3  are  loaded. 
Taking  moments  about  right  end,  we  have 
Positive  shear  in       3-5  =  iii  =  9(^  +  f  +  f  + 1)  =  15  tons. 
Negative  shear  in     3-5  =  iJ,  -  9  =  9  x  f  -  9  =  7.5  -  9 

=  —  1.5  tons, 
or,  negative  shear  in  3-5  =  -  9  x  ^  =  -  1.6  tons,  as  before. 

Likewise, 

5-7  =  1(1 +  2 +  3)  =  9  tons. 
5-7  =  - 1  (1  +  2)  =  -  4.5  tons. 
7-9  =  f  (1  +  2)  =  4.5  tons. 
7-9  =:  -  f  (1  +  2  +  3)  =  -  9.0  tons. 
9-11  =  I  X  1  =  1.5  tons. 
Negative  shear  in    9-11  =  -  f  (1  +  2  +  3  -f  4)  =  - 15  tons. 
Negative  shear  in  11-13  =  -f  (14-2  +  3  +  4  +  5) 

=  -  22.6  tons. 


Positive  shear  in 
No;^ative  shear  in 
Positive  shear  in 
Negative  shear  in 
Positive  shear  in 


I  I 


■^^ 


the  shear 
•ad,  and  is 
)ft  of  this 
Hence  the 
'le  live  load 

ive  shears 
is  9  tons, 
sitive  live 
the  right 
I.  Hence, 
lly  loaded. 

iS. 

ire  loaded. 

:  15  tons. 
.5-9 

IS  before. 


IS. 


0  tons. 


- 16  tons. 


•■'i 


t 


■|f* 


lililDGE  TIlUf^SKS. 


69 


There  is  no  positive  shear  in  11-l.S. 

Tlie  greatest  positive  shears  are  ecpial  and  symmetric^al  to 
the  greatest  negative  shears.  Thus,  the  positive  shear  in 
l-;i  is  equal  numerically  to  the  negative  shear  in  11-1.'J; 
the  positive  shear  in  3-5  is  equal  numerically  to  the  nega- 
tive shear  in  9-11 ;  and  so  on. 

Prob.  BO.  Find  the  greatest  positive  and  negative  shears 
for  each  panel  of  Fig.  25  when  the  apex  live  load  is  G  tons. 

Ans.  Positive  shears  =  21.0,  15.75,  11.25,  7.5,  4.5,  2.25, 
0.75,  0.0  tons. 

Negative  shears  =  0.0,  0.75,  2.25,  4.6,  7.5,  11.25,  15.75, 

21.0  tons. 

Prob.  51.  Find  the  maximum  positive  and  negative  shears 
for  each  panel  of  Fig.  28  when  the  apex  dead  and  live  loads 
are  5  and  9  tons,  respectively,  as  in  Prob.  47. 

Ans. 


Memiierh. 

2-4 

'■-e 

fi-S 

8  10 

10-12 

12-14 

Dead  load  shear 

Llvd  load  positive  shear     .    . 
Live  load  noffatlve  ahesr    .    . 

+27.60 

+4!)..')0 

0.00 

+22.60 
+41.25 
-  0.75 

+17.60 
+33.75 
-  2.25 

+  12.50 

+'i7.00 
-  i.M 

+  7.50 
+21.00 
-  7.60 

+  2.60 
+1.5.75 
-11.25 

Maximum  shear 

Minimum  shear 

+77.00 
.^.27.50 

+03.76 
+21.75 

+51.25 
+  1.5.25 

+TO.r)0 
+  s.oo 

+28.60 
0.00 

+18.26 
-  8.76 

The  maximum  shear  is  always  positive  in  the  left  half  of 
the  truss,  but  the  minimum  may  be  either  positive  or  nega- 
tive. In  this  problem  we  see  that  a  negative  shear  can 
occur  in  panel  12-14,  and  that  there  is  no  shear  in  panel 
10-12,  while  the  shears  in  all  the  other  ppnels  are  positive. 
Members  to  the  left  of  10,  and  of  course  in  the  four  right 
hand  panels  also,  are  therefore  subjected  to  but  one  kind  of 
stress,  while  in  the  four  middle  panels  they  may  be  sub- 


60 


UOOFS  AND  nnwGES. 


jected  to  either  kind  of  stress,  and  hence  must  be  counter- 
braced.  Though  there  is  no  negative  shear  in  panel  10-12, 
it  would  be  couuterbraced  for  safety. 

Art.  23.  The  Warren  Truss.  —  The  chord  stresses  in 
the  Warren  Truss  nuiy  be  couiputeil  by  the  method  of  chord 
increments,  (2)  of  Art.  19,  and  the  web  stresses  by  the 
method  of  Arts.  18  and  22.  The  dead  and  live  load  stresses 
may  be  found  separately  and  their  sum  taken  for  the  maxi- 
mum and  minimum  stresses;  or  the  maximum  and  mini- 
mum stresses  of  any  member  may  be  determined  directly, 
from  a  single  equation,  by  placing  the  dead  and  live  loads 
in  proper  position.  In  tlie  following  solution  of  the  deck 
Warren  truss  the  dead  and  live  load  stresses  are  found 
separately. 

I*rob.  52.  A  deck  Warren  truss,  as  a  deck  railroad  bridge. 
Fig.  29,  has  10  panels,  each  12  feet  'long,  its  web  members 

7       0      It      11'     9'      f     S'       S' 1 


1   S       !''       7       "      '•      11       'J        I       !>        J    f 

AAAAA/\/\/\/\/\ 

t     h     6     s     10    a    10'  s     0     If     ' 


Kig.  so 

all  forming  equilateral  triangles  (Art.  14) ;  the  dead  load  is 
given  by  formula  (2),  Art.  15,  the  live  load  is  1500  lbs.  per 
foot  per  truss :  find  the  maximum  and  minimum  stresses  in 
all  the  members. 
The  dead  panel  load  per  truss  from  formula  (2) 

_  ^7^020  +  600)12  ^  8g4o  j^s.  =  4.32  tons 
<> 

=  say,  4  tons,  for  convenience  in  computation. 
Live  panel  load  per  truss  =  -^^^jjjjjj —  =  9  to^^s- 


•A. 


% 


1 


5  counter- 
lel  10-12, 

itresses  in 
I  of  chord 
53  by  the 
id  stresses 
the  raaxi- 
and  mini- 
l  directly, 
live  loads 
the  deck 
are  found 

ad  bridge, 
>  members 


ead  load  is 
DO  lbs.  per 
stresses  in 

) 

ans 

ition. 


BlilDUE  TltUSfiEH. 


61 


Since  the  loads  are  distributed  uniformly  along  the  joints 
of  the  upper  chord,  .3  and  3'  will  receive  three  fourths  of  a 
panel  load  each.  The  dead  and  live  load  stresses  will  be 
fou:id  separately,    tan  6  =  .577 ;  sec  d  =  1.1547. 

Dead  Load  Stresses. 

Left  reaction  =  4.75  x  4  =  19  tons.  This  is  also  the  shear 
in  panel  1-3.  The  dead  load  shear  in  each  of  the  other 
panels  is  found  by  subtracting  the  loads  on  the  left  of  the 
panel  from  the  abutment  reaction.  We  have  then  the  fol- 
lowing chord  stresses  by  chord  increments : 

Upper  Chord  Stresses. 

Stress  in     3-5  =  (19  +  16)  X  .577  =  -  20.20  tons. 

Stress  in     5-7  =  (19  +  2  x  16  -1- 12)  x  .677  =  -  36.36  tons. 
Stress  in     7-9  =  (19  +  2  x  16  4-  2  X  12  +  8)  x  .577 

=  -  47.90  tons. 
Stress  in  9-11  =  (75  +  2  X  8  +  4)  X  .577  =-54.82  tons. 
Stress  in  11-11'=  (91  +  2  x  4)  X  .577  =  -  57.12  tons. 

Stress  in      1-3  =     00.00  tons. 

Lower  Chord  Stresses. 

Stress  in      2-4  =  19  x  .577  =  10.96  tons. 

Stress  in  4-6  =  (19  +  2  x  16)  x  .577  =  29.42  tons. 
Stress  in  6-8  =  (51  +  2  X  12)  x  .577  =  43.28  tons. 
Stress  in  8-10  =  (75  +  2  x  8)  x  .577  =  52.51  tons. 
Stress  in  10-12  =  (91  +  2  x  4)  x  .577    =  57.12  tons. 


-»-«wBnw!wggS';' ■ 


ipi 


6*! 


ROOFH  AM)  nuiDaiis. 


Weh  Stuhsseh  (iiY  Akt.  18). 

Stress  in      2-3 = 1 1)  x  1 . 1  '>i  =  -  21 .85  tons. 
Stress  in      3-4=10  x  1.154:^^  +18.40  tons=  -stress  in  4-5. 
Stress  in      5-6=12x1.154= +13.84  tons= -stress  in  L    ". 
Stress  in      7-8=  8x1.154=+  9.23  tons  =- stress  iu  8-9. 
Stress  iu    9-10=  4x1.154=+  4.G1  tons 

=  —stress  in  10-11. 
Stress  in  11-12=0.00  =  -stress  in  12-12'. 

Stress  in     1-2=^  of  a  panel  load=l  ton. 


Live  Load  Stresses. 

The  maximum  live  load  chord  stresses  occur  when  the 
live  load  covers  the  whole  length  of  the  truss  (Art.  20); 
hence  they  are  found  in  precisely  the  same  way  as  the  dead 
load  chord  stresses  above;  or  we  may  find  them  by  multi- 
plying the  above  dead  load  chord  stresses  by  the  ratio  of 
live  to  dead  panel  load,  which  =  2^  here,  giving  us  the 
following  chord  stresses : 


Uppke  Chord  Stresses. 

1-3  =  00.00  tons. 
3-5  =  —  45.45  tons. 
5-7  =  -  81.81  tons. 
7-9  =  -  107.77  tons. 
9-11  =  - 123.34  tons. 
U-ll'=- 128.52  tons. 


Lower  Chord  Stresses. 

2-4  =   24.66  tons. 

4-6=    66.19  tons. 

6-8=   97.38  tons. 

8-10  =  118.15  tons. 

10-12  =  128.52  tons. 


BRIDGE  TKUSSES. 


m 


ss  in  4-5. 

S3  in  t   r. 
88  iu  8-9. 


when  the 

(Art.  20); 
s  the  (lead 
by  multi- 
lie  ratio  of 
ng  us  the 


•  tons. 
>  tons. 
i  tons. 
)  tons. 
]  tons. 


Wkh  Stkk.sses. 

The  stress  in  any  web  member  is  equal  to  the  shear  in 
the  section  whicli  cuts  that  member  and  two  liorizontal  chord 
members,  multiplied  by  the  secant  of  tlie  angle  which  the 
web  member  makes  with  the  vertical  (Art.  18) ;  the  stress 
is  therefore  a  maximum  when  the  shear  is  a  maximum. 
The  maximum  positive  live  load  shear  in  any  panel  occurs 
when  all  joints  on  the  right  are  loaded  and  the  joints  on 
the  left  are  unloaded  (Art.  22). 

The  maximum  jjositive  live  load  shear  for  2-3  will  occur 
when  the  truss  is  fully  loaded.     We  have  then 

Stress  in  2-3  =  9  X  4.76  x  1.154  =  -  49.30  tons. 

The  maximum  positive  shear  for  3^  and  4-5  will  occur 
when  all  joints  except  3  are  loaded  (Art.  22).  Taking 
moments  about  the  right  end,  and  multiplying  by  sec  6,  we 
have 


Stress  in  3-4 
=  A(f  +  3-f- 
=  Ax^|^x 

Stress  in  5-6 
=  /Tra  +  3  + 
=  33.13  tons : 

Stress  in      7-8 


5  +  7  4-  9  + 11  + 13  + 15  -I- 17)  X  1.1547 
1.1547  =  41.96  tons  —  —  stress  in  4-5. 

5  +  7  +  9  + 11  + 13  -!- 15)  X  1.1547 
=  —  stress  in  6-7. 

=  Ad +  3 +  5+7+9+11  +  13)  X  1.1547 
=  25.33  tons  =  —  stress  in  8-9. 


Stress  in    9-10  =  ,_,\  (|  +  .3  +  5  +  7  +  9  +  1 1)  x  1.1547 
=  18.58  tons  =  -  stress  in  10-11. 

Stress  in  11-12  =  /^(J  +  3  +  5  +  7  +  9)  x  1.1547 
=  12.87  tons  =  -  stress  iu  12-11'. 


:'>'«ite»itti^y«>iii«aj@^tBa!8W»MHeffii«K«r^ 


^ 


04  HOOFS  AND   IIIIIDGKS. 

Stress  in  ll'-lO'  =  /o  (1  +  -^  +  •'>  +  ")  x  l.ir)47 

=  8.11)  tons  =  -  stress  in  lO'-U'. 

Stress  in      9'-8'  =  .'V  (3  +  .*i  +  5)  x  1.1517 

=  4.55  tons  =  —  stress  in  8'-7'. 

Stress  in      7'-G'  =  ^'V  (f  +  •^)  ^  l-^''*-*^ 

=  l.DC  tons  =  —  stress  in  (5-6'. 

Stress  in      5'-4'  =  ^\  (|)  x  1.1547 

=  0.40  tons  =  —  stress  in  4'-3'. 

Stress  in      3'-2'  =  0.00  tons. 

Collecting  the  above  results  we  may  enter  them  in 
as  follows : 

TABLE  OF  STRESSES  IN  ONE  TKUSS. 
Upper  Chord  Stuessbs. 


a  table 


Memhihs. 

1-3 

8-5 

5-7 

7-» 

1)  11 

11-11' 

Dead  load 
Live  load 

00.00 
00.00 

-20.20 
-46.45 

-  30.30 

-  81.81 

-  47.00 
-107.77 

-  54.82 
-123.34 

-  57.12 
-128.52 

-185.04 

-  67.12 

Max.  stress 
Mill,  stress 

00.00 
00.00 

-05.05 
-20.20 

-118.17 
-  30.30 

-165.07 
-  47.00 

-178.10 

-   54.82 

Lower  Chord  Stresses. 


Mkmiieks. 

i-'-i 

4-6 

C-S                  8-10 

10-12 

Dead  load    .  . 
Live  load  .  .  . 

+  10.0(J 
+  24.C,0 

+29.42 
+  00.19 

+  43.28 
+  97.:5H 

+  52.51 
+  118,15 

+  57.12 
+  128.62 

Max.  stress .  . 
Min.  stress  .  . 

+  35.02 
+  10.00 

+95.01 
+  29.42 

+  140.00 
+  43.28 

+  170.00 
+  62.51 

+  186.64 
+  67.12 

-y. 

V. 
5'. 
3'. 

)m  in  a  table 

3S. 


I  a 


11 

11-11' 

4.82 
3.34 

-  57.12 
-128.52 

8.10 

)4.82 

-185.04 
-  67.12 

) 

10-12 

.51 
15 

+  57.12 
+  128.62 

.0(i 
.51 

+  186.64 
+  57.12 

BlilDQE  TliU88E8. 


66 


Weu  Stresses. 

Mkmhkkn. 

2-3 

it  4 

4-5 

frfl 

8-7 

^  Dead  load    .... 
J  j  From  rif?lit  .  .  . 
i  (  From  left .  .  .  . 

-21.8.1 

-40,;to 

00.00 

+  18.40 
+  41.110 
-  0.40 

-18.40 
-41.00 
+  0.40 

+  13.84 
+  33.13 
-  1.90 

-13.84 
-33.13 
+   1.90 

Max.  8tri!S8 

Mill,  stresa 

-71.15 
-21.86 

+  60.42 
+  18.00 

-00.42 
- 18.00 

+40.07 
+  11.88 

-46.07 
-11.88 

M  KM  HERS. 

T-9 

8-9 

9-10 

10-11 

11-12 

^  Dead  load   .... 
§  j  From  right  .  .  . 
1  1  From  left .... 

+  9.23 
+25.33 
-  4.56 

-  9.23 
-26.33 
+  4.55 

+  4.01 
+  18.58 
-  8.10 

-  4.01 
-18.68 
+  8.10 

+  0.00 
+  12.87 
-12.87 

Max.  stress 

Min.  stress 

+  34.66 
+  4.08 

-34..'i0 

-  4.08 

+23.19 
-  3.58 

-23.19 
+  3.58 

+  12.87 
-12.87 

We  see  from  this  table  of  tueb  stresses  that,  (1)  when  the 
live  load  comes  on  from  the  right,  all  the  web  members  in 
the  left  half  of  the  truss  are  subjected  to  but  one  kind  of 
stress,  that  is,  the  members  2-3,  4-5,  6-7,  8-9,  10-11  are 
subjected  to  compressive  stresses,  and  the  members  3-4,  5-6, 
7-8,  9-10,  11-12  are  subjected  to  tensile  stresses ;  (2)  when 
the  live  load  comes  on  from  the  left,  all  the  web  members 
to  the  left  of  9-10  are  subjected  to  but  one  kind  of  stress, 
but  the  members  9-10,  10-11,  11-12  have  their  stresses 
changed,  that  is,  the  stresses  in  9-10  and  11-12  are  changed 
from  tensile  to  compressive,  while  that  in  10-11  is  changed 
from  compressive  to  tensile. 

Hence,  the  members  2-3,  4-5,  6-7,  8-9,  should  be  struts 
to  carry  compression  only,  and  the  members  3-4,  6-6,  7-8, 


'iiaxiiBjaiWMia 


I? 


i 


liOUFS   AM)    ltl!lJ)(li:S. 


■■■, 


!i 


It  !i 


should  be  tit^s  to  carry  tension  only,  while  9-10,  10-11, 
11-lL'  shouUl  lie  nionibcrH  I'apabli'  of  iv,sistin(<  both  coni- 
l)rcs,sion  and  tcn.siini,  and  should  thnieforo  be  amiiter  hritcea 
(Art.  1):  and  the  same  is  true  for  the  ri.Ljht  half  of  the  truss. 

In  this  solution  for  web  stresses  the  live  load  was  brought 
on  from  the  n'(jlit,  and  the  niaxiuuna  /losiUce  shear  was 
found  in  eaeh  pantd  of  the  rifiht  half  of  the  truss,  and  then 
the  resulting,'  stress ;  if  preferred,  the  live  load  may  \w 
brought  on  from  the  lif},  and  the  maximum  wyative  shear 
bo  found  in  each  iianel  of  the  Jeft  half  of  the  truss  (Art.  22), 
and  then  the  resulting  stress.  Thus,  the  maximum  jjositive 
shear  in  any  paii'l  in  the  right  half  of  the  truss  as  (J'-8'  has 
the  same  numerical  value  as  the  maximum  negative  shear 
in  the  corresponding  lueiubor  C-8  in  the  left  half;  and  the 
resulting  stress  in  any  member  7-8'  lias  the  same  value  as 
that  in  the  corresponding  nuMuber  7-8. 

The  maximum  and  miniminn  stresses  may  be  determined 
directly  from  a  single  equation  by  placing  the  dead  and 
live  loads  in  proper  position. 

Thus,  for  the  nuiximum  stress  in  8-9,  wo  pass  a  section 
cutting  it,  place  the  live  load  on  the  right,  and  have 

Max.  stress  in  8-9==-[2x4-f /^(|+3+5  +  7+9+ll+13)] 
X  1.1547=  —34.57  tons,  as  before. 

For  the  niinimuni  stress  tlie  live  load  is  reversed  and 
covers  the  truss  on  the  left  of  the  section.     Thus 

Min.  stress  in  8-9=-[2x4-^\(f+3+5)]xl.l547 
=  — 4.()9  tons,  as  before. 

Prob.  53.  A  deck  Warren  truss,  like  Fig.  29,  has  10 
panels,  each  10  feet  long,  its  web  mendiers  all  forming  equi- 
lateral triangles;  the  dead  load  is  800  lbs.  per  foot  per 
truss,  and  the  live  load  is  1000  lbs.  per  foot  per  truss ;  the 


f'^f' 


..i 


Bnil'HE  TRUSSEfi. 


fi7 


9-10,  10-11, 

\\;  both  coiu- 
•oitnter  hraccti 
I  of  the  truss. 
I  was  hiDUglit 
[•e  slioiar  was 
US8,  and  then 
load  may  he 
wyutivc  shear 
•uss  (Art.  22), 
iiuini  i)()sitive 
s  as  (5-8'  has 
egative  shear 
lalf ;  and  the 
same  vahie  as 

le  determined 
he  dead  and 

ass  a  section 
have 

-9+11  +  13)] 
1  before, 
reversed  and 

lUS 

<  1.1547 


;.  20,  has  10 
foriiiin^  equi- 
per  foot  i)Pr 
er  truss:  the 


■^.'i.  J5^ 


joints  3  and  3'  receive  three  fourths  of  a  i)anol  h>ad  each: 
iind  the  maximum  and  minimum  strosHi".-)  in  M  the  members. 


At)s, 


Ui-i'Kit  CiioKi>  Sthksse 


Mkmhkrh. 

1  a 

8-5 



-60.60 
-20.20 

r>7 

TO 

9-11 

11-11' 

Max,  Ht  n'H.st'8 
iMin.Mtri'KNr.s 

00  00 
00.00 

-  100.08 

;iO.:j« 

- 14:(.70 

-  47.02 

-  ltt4..V.' 

-  64.84 

-17I.4M 

-  .17.1(1 

LOWKK 

!^iioiti)  .Strksses. 

MKMIIKR!), 

9-» 

4-a 

0  9 

f-in 

KI-12 

Max.  stresses  .  .  . 
Mill,  HtrcsHPS    .   ,   . 

+32.88 
1  10.06 

+  88.20 
+  29.40 

+  120.8.4 
+  4;J.28 

+  1.')7.50 
+  62.52 

+  171.48 
+   57. 16 

Web  Strbsses. 


Mkuiikkh. 

li-3 

8-4 

4-8 

6-0 

6-1 

Max.  8trt's.se8  .  .  . 
Mill.  stroascH   .  .  . 

-65..52 

-21.H4 

+  65.80 
+  18.12 

-65.80 
-18.12 

+  r.i.2!i 
+  i2.o;i 

-4.'5.20 
-12.00 

MKMIIIiRK. 

7-8 

8-9 

9-10 

i«-ii 

u-ia 

Max.  strcsaes  .  .  . 
Mill,  stresses    .  .  . 

+;{1.78 
+  5.18 

-31.78 
-  6.18 

+21.ia 
-  2.70 

-21.1.3 
+  2.70 

+  11.46 
-11.46 

Prob.  54.  A  deck  Warren  truss,  like  Fi<?.  29,  has  8  panels, 
each  15  feet  long,  its  web  members  all  forming  equilateral 
triangles;  the  dead  load  is  given  by  formula  (2)  of  Art.  15, 
the  live  load  is  1000  lbs.  per  foot  per  truss,  tlie  joints  3  and 
3'  receive  three  fourths  of  a  panel  load  each  :  find  tlie  maxi- 
mum and  minimum  stresses  in  all  the  members. 


*i^v>w^,^«(^£gusaiSi^i^^^^sgi@l$a6^^»^aiaaiista9£^&tie^-^^'?MttiMfi»t@if»^  > 


]i(JOFS  AM)   nil  I  DUES. 


1   ■ 


Aug.  Max.  stresses  in  upimr  dionl 

=  0.00,  -()7.()8,  -117.97,  -UH.L'o,  -158.17  tons. 

Mill.  HtiTH.se.s  in  uppt'i'  chonl 

=  0.00,   -L'l.OO,    -.'50.01,     -Ki.OI,   -lO.O'J  tons. 

Max.  sti'CMsi's  in  lower  choid 

=  +  n7.r>s,  +07.00,  +i;w.i5,  -\  ir,HA~  touH. 

Min.  HtrcHscs  in  lower  cliord 

=  4-1 100,  +'M.W,  +'lli.87,  +40.00  tona. 

Max.  web  Htresses 

=  _  74.00,  +00.08,  -00.08,  +48.-1.3,  -  4.'U.'J,  +  27.57, 

-  '27.07,  + 1;{.08  tons. 

Min.  web  Htrcssea 

=  -U3.L'7,   +18.20,    -18.20,   +0.23,   -9.23,    -1.35, 
+  1..35,  - 13.08  tons. 

Pivb.  65.  A  throiiKli  Warren  truss,  Fij,'.  2(5,  has  10  panels, 
each  12  feet  loiif,',  its  web  nienilier.s  all  forniinK  eciuilateral 
triangles;  the  dead  load  is  500  lbs.  per  loot  per  truss,  and 
the  live  load  is  8.'i4  lbs.  per  foot  per  truss :  find  the  maxi- 
nulla  and  niiniiuum  stresses  in  all  the  members. 

Alts.    Upper  ehord  max. 

=  -41.52,  -73.92,  -90.90,  -110.80,  -115.44  tons. 

Ui»per  ehord  min. 

=  _15.57,  -27.72,  -3G.30,  -41.55,  -43.29  tons. 

Lower  chord  max. 

=  +20.80,  +57.70,  +85.44,  +103.92,  +113.12  tons. 

Lower  cliord  min. 

=  +7.80,  +  21.00,  +  32.04,  +  38.97,  +  42.42  tons. 

Web  stresses  max. 

=  -41.55,  +41.55,  -32.91,  +32.01,  -24.81,  +24.81, 

-  17.31,  + 17.31,  -  10.37,  +  10.37  tons. 


%t^f 


m^^'- 


ITtona. 
9  tuns. 

>ltH. 

.'J.I.'J,  +  27.57, 

9.23,    -  1.35, 

has  10  i)aiipls, 
1^,'  cciuilatcral 
per  trus.s,  and 
intl  the  maxi- 

s. 

.16.44  tons. 
1.29  tons. 
113.12  tons. 
B.42  tons. 

!4.81,  +24.81, 

s. 


'W^^^- 


UUWUE  Tit  U SUES. 


ti'J 


Web  sticsses  niin. 

=:_  ir,.57,  +  15.57,  -11.55,  +11.55,  -(5.91,  +0.91, 

-  1.73,  +  1.73,  +  4.()('.,  -  4.0(5  tons. 

Prob.  66.  A  through  Wanen  tni,ss,  like  Kig.  2fi,  lias  K 
panels,  each  15  feet  long,  its  web  nuMiibt'is  all  turniing  equi- 
lateral triangles;  the  dead  load  is  SOO  lbs.  per  foot  per  truss, 
aiul  the  live  load  1(500  lbs.  per  fool  per  truss;  find  the  uiaxi- 
muni  and  luininuim  stresses  in  all  the  nieud)er8. 

Ans.  Upper  ehord  max. 

=  -48.48,  -  83.10,  -  103.92,  -  110.88  tons. 

Upper  chord  min. 

=  -  10.10,  -  27.72,  -  34.04,  -  30.90  tons. 
Lower  ehord  max. 

=  +  24.24,  +  05.70,  +  93,00,  +  107.40  tons. 

Lower  chord  min. 

=  +  «.()S,  +  21.92,  +  31.20,  +  35.80  tw.is. 

Web  stresses  max. 

=.-48.48,  +48.48,  -35.08,  +35.08,  -24.20,  +24.20, 

-  13.80,  +  13.80  tons. 

Web  stresses  min. 

=  -  10.10,  +  48.48,  -  10.40,  +  10.40,  -  3.48,  +  3.48, 
+  4.08,  -  4.08  tons. 

Prob.  57.  A  deck  Warren  truss,  like  Fig.  29,  has  7  ])anels, 
eacdi  15  feet  long  and  15  feet  deep,  its  web  nuMubers  all 
forming  isosceles  triangles;  the  dead  load  is  4  tons,  and  the 
live  load  is  9  tons,  per  panel  per  truss,  the  first  and  last 
joints  3  and  3'  receiving  three  cpiavters  of  a  panel  load  each, 
as  in  Vrobs.  51,  52,  and  53:  find  the  maximum  and  mini- 
mum stres.ses  in  all  the  members. 

Ans.  Upper  ehord  max. 

=  -  0.00,  -  37.38,  -  03.38,  -  70.38  tons. 


*.^i'!ft5&^3,!K^i3fclt3iil0S^5'Sft**?'''****^*^  ■*j*w»/st-'«*»*''waR*sss«wts*w.»«*efi»we«si9B^^ 


ROOFti  AND  liRIDGES. 

Upper  clionl  iiiin. 

=  _  ().(K),  _  11.50,  -  19.60,  -  23.50  tons. 
Lower  chord  max. 

=  4-  -1.1.'5,  +  5y,0i~,    ,-  TlJ.ia,  +  79.6.'}  tons. 
Lower  cliur'l  niiii. 

=  +  (i.-'ilS  +  10.50,  +  22.50,  +  24.50  tons. 
Web  stresses  max. 

=  -•47.2.".,  +;}(j.87,  -30.87,  +24.49,  -24.49,  +13.50, 
—  13.5(5  tons. 

Web  stresses  niin. 

=  -  14.63,   +  10.04,  -  10.64,   +  4.01,  -  4.01,  -  4.05, 
+  4.05  tons. 

Art.  24.  Mains  and  Counters.  —  In  the  Pratt  truss, 
Fig.  30,  all  the  verticals,  except  1-4  and  l'-4',  are  to  take 
compression  only,  and  all  the  inclined  members,  except  1-2, 

1       .1        5       5'       5'      1' 


1-2'  are  to  take  tension  only.  The  two  inclined  members, 
1-2  and  l'-2',  are  usually  called  the  "  inclined  end  posts." 
The  verticals  t-4  and  l'-4'  do  not  form  any  part  of  the 
truss  proper,  since  they  serve  only  to  carry  the  loads  at  4 
and  4'  np  to  the  hip  joints  1  and  1';  they  are  called  hip 
verticals. 

The  members  1-0,  3-8,  3-8',  and  I'-O'  are  mains,  or  main 
ties.  They  are  the  onlj^  inclined  ties  that  are  pnt  under 
stress  by  the  action  of  the  dead  load,  or  by  a  uniform  dead 
and  live  load  extending  over  the  whole  truss.  The  mem- 
bers 5-6,  5-8',  5'-8,  and  5'-6',  in  dotted  lines,  are  counters, 


"o^^^HE^^ 


BRIDGE  TliVSSES. 


71 


.49,  +  13.50, 

l.Ol,  -4.or>, 

Pratt  truss, 
,  are  to  take 
,  except  1-2, 


ed  members, 
I  end  posts." 
part  of  the 
e  loads  at  4 
e  called  hip 

tins,  or  main 
e  put  under 
uiform  dead 
The  mem- 
are  counters, 


or  counter  ties.  In  the  I'ratt  truss  there  are  no  l-ounter 
braces  (Art.  1).  These  four  counters  are  called  into  j)lay 
<mly  when  the  live  load  covers  a  part  of  the  truss. 

Thus,  if  a  load  be  placed  at  the  joint  (>',  it  is  carried  by 
the  truss  to  the  abutments  at  2  and  2'.  The  part  of  this 
load  whi(;h  goes  to  2  may  be  conceived  as  being  carried  up 
to  o',  down  to  8',  up  to  5,  down  to  8,  up  to  3,  down  to  o,  up  to 

1,  down  to  the  alnitment  at  2.  The  other  part  of  this  load, 
which  goes  to  the  right,  passes  up  to  1',  then  down  to  2'. 
For  this  loading,  the  counters  5 -G',  5-8',  and  the  mains  3-8, 
l-(5,  and  I'-O'  are  put  under  stress,  while  the  counters  5'-8, 
r>-{),  and  the  main  tie  3-8'  are  idle,  and  might  be  removed 
without  endangering  the  truss.  If  the  two  middle  joints, 
8  and  8',  are  loaded  equally,  the  part  of  the  load  at  8'  going 
to  2'  is  just  balanced  by  the  part  of  the  load  at  8  going  to 

2,  and  hence  there  is  no  stress  in  the  intermediate  web 
members  5-8',  5'-8,  5-8,  and  5'-8',  nor  in  the  counters  5-6 
and  5'-6'.  If  the  live  load  going  from  6'  to  the  left  abut- 
ment is  greater  than  th'>  lieaci  load  going  from  8'  to  the 
right  abutment,  the  countc;  5-6'  must  be  inserted,  if  this 
counter  5'-C'  were  not  inserted,  the  panel  G'-8'  would  be 
distorted.  The  load  at  G'  would  bring  the  opposite  corners 
3'  and  8'  nearer  together,  and  the  opposite  corners  5'  and 
G'  farther  apart,  because  the  main  tie  3'-8'  cannot  take  com- 
pression. Whenever  the  live  load  would  tend  to  cause  com- 
pression in  any  main  tie,  a  counter  must  be  inserted  uniting 
the  other  corners  of  the  panel.  Both  diagimals  in  any  panel 
cannot  have  compression  or  tension  at  the  same  time. 

If  the  action  of  the  dead  and  live  loads  tend  to  sul)ject  a 
member  to  stresses  of  op])osite  kiiuls,  the  resultant  stress 
in  the  member  will  be  equal  to  the  numerical  diffe-enec  of 
the  ojvposite  stresses,  and  will  be  of  the  same  kind  as  the 
greater.     Tims,  if  the  dead  load,  g(nng  from  8  to  the  left 


•■<»«-=«»f«*o»,;jaiSiSg,c^j;,l55|r,-.>nj,,rj.ij,.i 


r,v^siiift««-^/-*c>«.v~v.-»4is«(.i»v-'.  ■ 


tiS»i3«*^*«nHiayj;- 


72 


nOOFS  AND  nUlDGES. 


m 


t  I- 


I  1 


abutment,  shoukl  subject  the  main  tie  3-8  to  a  tension  of  8 
tons,  and  if  the  live  load,  going  from  0  to  the  right  abut- 
ment, should  tend  to  subject  the  same  tie  to  a  compression 
of  ;')  tons,  there  would  bo  a  resulting  tension  of  3  tons  in 
the  member  ,'J-8 ;  and  the  counter  5-0  would  not  be  needed 
for  this  loading.  Ihit,  if  the  dead  load  should  subject  the 
tie  3-8  to  a  tension  of  only  2  tons,  while  the  the  live  load 
should  tend  to  sidiject  the  same  piece  to  a  compression  of 
5  tons,  there  would  be  a  resulting  force  of  3  tons,  tending 
to  bring  the  joints  3  aiul  8  nearer  together,  or  to  separate 
the  joints  o  and  (j  from  each  other.  In  order  t.)  provide  for 
this  compression  or  tension,  either  a  brace  would  be  needed 
alongside  the  main  tie  3-8,  to  take  the  compression  of  3  tons, 
or  the  counter  tie  fl-G  would  be  needed  to  take  the  tension 
of  3  tons. 

In  the  Howe  truss.  Fig.  31,  the  verticals  are  to  take  ten- 
sion only,  and  the  inclined  members  are  to  take  compression 


Kit.'.  31 

only.  The  mains,  or  main  braces,  are  represented  by  the 
full  lines ;  the  dotted  lines  de  lote  counters,  or  counter 
braces.  Under  the  action  of  the  dead  load,  or  of  a  uniform 
dead  and  live  load  covering  the  whole  truss,  the  main 
braces  are  the  only  diagonals  that  are  put  under  stress. 
The  counter  braces  are  called  into  play  only  when  the  live 
load  covers  a  part  of  the  truss,  as  iu  the  case  of  the  Pr  itt 
truss. 

Thus,  if  a  load  be  placed  at  the  joint  G,  the  part  of  it 
which  goes  to  the  right  abutment  L"  may  be  conceived  as 


|isl'#,^*^6Sfesv 


tension  of  8 
i  right  abiit- 
compression 
of  3  tons  iu 
[)t  be  needed 
I  subject  the 
bhe  live  load 
inpression  of 
tons,  tending 
f  to  separate 
)  provide  for 
Id  be  needed 
ion  of  3  tons, 
J  the  tension 

to  take  ten- 
I  compression 


3nted  by  the 
I,  or  counter 
of  a  uniform 
3S,  the  main 
under  stress, 
fhen  the  live 
of  the  I'r.tt 

he  part  of  it 
conceived  as 


niilDGE  riiUSSES. 


78 


being  carried  up  to  3,  down  to  8,  up  to  5,  down  to  8',  up  to 
5',  down  to  G',  up  to  3',  i;owu  to  4',  up  to  1',  down  to  2'. 
If  the  live  load  going  from  6  to  the  right  abutment  is 
greater  than  the  dead  load  going  from  8  to  the  left  abut- 
ment, the  counter  brace  3-8  must  be  inserted.  If  this 
counter  3-8  were  not  inserted,  the  panel  6-8  would  be  dis- 
torted. The  load  at  G  would  bring  the  op])osite  corners  3 
and  8  nearer  together,  and  the  opposite  corners  6  and  6 
farther  apart,  because  the  main  brace  5-0  cannot  take 
tension.  But  if  the  live  load  going  from  G  to  the  right 
abutment  is  less  than  the  dead  load  going  from  8  to  the  left 
abutment,  the  counter  brace  3-8  will  not  be  needed  for  this 
loading.  The  main  and  counter  'n  any  panel  cannot  boih 
take  stress  at  the  same  time  by  any  system  of  loading. 

We  may  determine  where  the  counters  are  to  begin,  as 
follows:  Consider  the  left  half  of  the  truss,  and  let  the 
live  load  come  on  from  the  left ;  then  the  dead  and  live  load 
shears,  or  stresses,  are  of  opposite  kinds.  It  results  from 
this  at  once  that  counters  —  counter  ties  in  the  Pratt,  and 
counter  braces  in  the  Howe  —  must  begin  in  that  panel  in 
which  the  stress  or  shear  caused  by  the  live  load  is  greater 
than  that  of  the  opposite  kind  caused  by  the  dead  load. 

Thus,  in  Fig.  30,  let  the  dead  panel  load  be  3  tons  and 
the  live  panel  load  14  tons.  Then  we  have  the  following 
maximum  negative  shears : 


Shear 

in  2-4 

=  4-9- 

-0 

=  +  9. 

Shear 

in  4-G 

=  4-6- 

-1 

xl4 

=  4-4. 

Shear 

in  6-8 

=  4-3- 

-14 

(1  + 

f) 

=  -3. 

Hence  the  counters  must  begin  in  the  third  panel,  and 
therefore  the  third,  fourth,  and  fifth  panels  must  have 
counter  ties  for  this  loading. 


■'  -■•««'**«w;«»i«»SK*«AV»*;#««R'iT^<**t'*^Jft"MaimE?t.MBa«!^^ 


74 


UOOFS  AND  niilDGES. 


Prol).  58.  Tn  the  Howe  truss  of  Fig.  28,  with  12  panels, 
the  (h'iul  [)iUiol  h);i(l  liciiij,'  4  tons  ami  tlic  live  1)  tons,  find 
the  unnil)ei'  of  panels  to  he  conuterbracetl. 

Alls.  The  5th,  Gth,  7th,  and  8th  panels  must  have  counter 
braces. 

Art.  25.  The  Howe  Truss.  —  The  niaxinnini  and  mini- 
mum stresses,  both  in  the  chords  and  the  web  menibers  of 
the  Howe  truss,  may  now  be  computed  by  the  principles  of 
Arts.  18,  19,  22,  and  24.  The  dead  and  live  load  stresses 
may  be  found  separately,  and  their  sum  taken  for  the  maxi- 
mum and  minimum  stresses;  or,  the  maximum  and  mini- 
mum stresses  of  any  inember  may  be  determined  directly, 
from  a  single  equation,  by  placing  the  dead  and  live  loads 
:  ptoper  position.  In  the  following  solution  of  the  through 
ilowe  t"uss,  tliis  method  is  the  one  employed : 

J^rub.  59.    A  through  Howe  truss,  as  a  through  railroad 

bridge,  Fig.  32,  has  10  panels,  each  12  feet  long  and  12  feet 

3 5 7         .1        ;;        fi'        ?'____   5' s' 


i'l;;.  33 


deei);  the  dead  load  is  given  by  formula  (4),  Art.  15,  the 

live  load  is  loO(»  lbs.  per  foot  per  truss:  liud  tlie  maxiintua 

and  mininuini  stresses  in  all  t)ie  members. 

From  tormula  (4),  Art.  15,  we  find  the  dead  panel  load 

per  truss 

=  (0.5  X J20  ±i!mi2  ^  3^30  1^^  ^  4  3^5  ^^^^ 

2 

=  say,  4  tons,  for  convenience  of  computation. 

T  •  111         4.  l''>00  X  12      Q  . 

Live  panel  load  ]»er  truss  =  — ^,7-;: —  =  ^  tons. 

tan0=  1,  sec  ^  =  1.41. 


lUv 


Bam 


itli  12  panels, 
e  1)  tons,  find 

;  have  counter 

until  and  niini- 

ib  ineiiibers  of 

iJiiiiciples  of 

load  stresses 

for  the  iiiaxi- 

nni  and  niini- 

liiied  directly, 

ind  live  loads 

of  the  through 

ongh  railroad 
iig  and  12  feet 

S' 


:),  Art.  15,  the 
the  maximtuii 

id  panel  load 

505  tons 

itation. 

ous. 


[.% 


L 


nUIDGE  TRUSSES. 


76 


Tlic  niaxinunn  and  iniiiiinnni  chord  ntresnes  may  he  written 
mit  at  once  (Arts.  I'.)  and  21).     They  are  as  follows: 

Max.  stresses  in  lower  chord  =  58.5,  104.0,  l;j().5,  150.0, 
102.5  tons. 

Min.  .stresses  in  lower  chords  18.0,32.0,42.0,48.0,50.0  tons. 

The  upper  cliord  stresses  may  be  written  from  the  lower 
at  once.  Thus,  stress  in  3-5  =  —  stress  iu  2-4 ;  and 
so  on. 

Weu  Stresses. 

The  maximum  stress  in  any  web  member  is  equal  to  the 
maximum  shear  in  the  section  which  cuts  that  member  and 
two  horizontal  (diord  members,  multiplied  by  the  secant  of 
the  angle  which  the  Aveb  member  makes  with  tlie  vertical 
(Art.  18).  Tlie  maximum  positive  live  load  shear  in  any 
panel  in  the  Icfi.  half  of  the  truss  occurs  when  all  joints  on 
the  right  an;  loaded  and  the  joints  on  thc^  left  are  unloaded 
(Art.  22).  For  the  maximum  stress  in  2-3,  we  pass  a  sec- 
tion cutting  it,  i»lace  the  live  load  on  the  right,  in  this  case 
covering  the  whole  truss,  and  have 

Max.  stress  in  2-3 
=  -  [4  X  4.6  +  ^!V(1  +  2  +  3  +  4  +5  +  6  +  7+  8  +  9)]  1.41 
=  -  82.48  tons. 

For  the  max.  stress  in  4-6  all  the  joints  on  the  right  of 
4  are  loaded.     Hence, 

Max.  stress  in  4-5 
=  -[4x3.5+V'(i(l  +  2+3+44- 

Max.  stress  iu  0-7 
=  -[4x2.5  + J-o  (1  +  2  +  3  4-. 

Max.  stress  in  8-9 
=  -[4xl.5  +  -f-a-(l+2  +  3+. 

Max.  stress  in  10-11 
=  -  [4  X  0.5  +  T-9^ (1  +  2  +  ...  +  5)]  1.41 


.+8)]  1.41=- -05.42  tons. 
+  7)]  1.41=- 49.03  tora. 
+  0)]  1.41= -35.10  tons. 

21.85  tons. 


I 


|*¥«Mi?i^;SBfo''»iAtf**'4«m»«--<B!**«jWn  *( 


^■■i* 


I  I 


76 


HOOFS  AND  mi  I  DOES. 


1 


■  f ' 


•mfi. 


For  the  minimum  str*'S3  in  any  membtn-  in  the  left  half 
of  thti  tniHs,  till'  live  load  i^  reversed  and  eovers  the  truss 
on  the  left  of  the  section  {Art.  2'2).     Hence, 

Min.  stress  in  2-3=  -[4  x  4.5  -  0]  1.41  =  -  25.38  tons. 

Min.  stress  in  4-5=  -  [4  x  3.5 -/^  x  1]  1.41  =  -18.47  tons. 

Min.  stress  in  G-7 

=  -  [4  X  2.5  -  ^^{l  +  2)]  1.41  =  -  10.29  tons. 

Min.  stress  in  8-9 

=  -  [4  X  1.5  -  -,9^  (1+  2  +  3)]  1.41  =  -  0.85  tons. 

Min.  stress  in  10-11 

=  _  [4  X  0.5  -  ^85  (1  +  2  +  3  +  4)]  1.41  =  +  9.87  tons. 

That  is,  the  minimum  stress  in  10-11  is  a  tension  of  9.87 
tons.  But  as  the  diagonals  in  the  Howe  truss  are  to  be  sub- 
jected only  to  compression,  the  brace  10-11  cannot  take  ten- 
si(jn,  the  counter  brace  9-12  must  therefore  be  inserted  to 
take  this  9.87  tons,  which  is  the  excess  of  that  part  of  the 
live  load  goint?  from  8  to  the  right  abutment  over  that  part 
of  the  dead  load  going  from  10  to  the  left  abutment.  If  a 
tie  were  placed  alongside  the  main  brace  10-11,  it  would 
take  this  tension  of  9.87  tons,  and  the  counter  brace  9-12 
would  not  be  needed;  or,  if  the  diagonal  10-11  were  built 
so  as  to  take  either  tension  or  compression,  the  coimter 
9-12  would  not  be  needed.  In  this  latter  case  the  member 
10-11  would  have  a  range  of  stress  from  —  21.85  tons  to 
-I-  9.87  ton.s,  the  former  when  the  live  load  covered  all  the 
joints  on  the  right  of  it,  the  latter  when  it  covered  all  the 
joints  on  the  left.  But  as  the  diagonals  are  to  be  subjocited 
only  to  compression  in  this  truss,  the  minimum  stiess  in 
10-11  is  zero,  and  the  tension  of  9.87  tons  in  10-11  becomes 
the  maximum  compression  iii  the  counter  brace  9-12.  The 
three  diagonals,  4-5,  G-7,  8-9,  are  always  in  compression. 


'  igSii3iSsiiMSitfcsgs^-*s*t' 


— -v^-at^tt  i"*!^^. 


Ii!iIJ)GE  riiUSSES. 


77 


the  left  half 
era  the  truss 

25.38  tons. 
::— 18.47  tons. 

I  tons. 

).85  tons. 

+  9.87  tons. 

msion  of  9.87 
are  to  he  suh- 
iiiiot  take  ten- 
ae  inserted  to 
,t  part  of  the 
iver  that  part 
iitment.  If  a 
1-11,  it  would 
pv  brace  9-12 
■11  were  built 
,  the  connter 
le  the  member 
21.8r>  tons  to 
ivered  all  the 
)vered  all  the 
J  be  subjocted 
aum  St  1  ess  in 
lO-1  1  becomes 
m  9-12.  The 
coiiipressiou. 


Hence,  theoretically,  there  is  only  one  counter  brace  needed 
in  the  h'ft  half  of  this  truss;  but,  practically,  a  counter 
brace  would  be  put  in  the  fourth  panel  also,  so  that  the 
tniss  would  have  counter  braces  in  its  four  middle  panels. 

The  maximuui  stress  in  any  vertical,  except  the  middle 
one,  is  e(pial  to  the  nuiximum  positive  shear  in  the  section 
cutting  it  and  two  chord  members,  or  it  is  equal  tt>  the 
maximum  positive  shear  in  the  panel  toward  the  abutment 
from  the  vertical.     Thus, 

Mi  X.  stress  in  3-4  =  13  X  4.5  =  58.5  tons. 

Max.  stress  in  5-() 

=  [4  X  3.5  +  .9(1  +  2  +  •••  +  8)j  =  4G.4  tons. 

Max.  stress  in  7-8 

=  [4  X  2.5  +  .9  (1  +  2  +  -  +  7)]  =  ao.a  tnna. 

Max.  stress  in  9-10 

=  [4  X  1.5  +  .9(1  +  2  +  •••  +  6)]  =  24.9  tons. 

The  maximum  stress  in  the  middle  vertical  11-12  is  equal, 
either  to  the  maximum  positive  shear  in  panel  10-12,  or  to 
a  full  panel  dead  and  live  load,  whichever  is  the  greater. 
For,  if  this  maximum  positive  shear  is  greater  than  a  panel 
load,  then  the  shear  in  the  next  right  hand  panel  under  the 
same  loading  is  also  positive,  and  the  counter  Imice  in  that 
panel  will  he  in  action,  thus  making  the  shear  in  panel  10-12 
the  same  as  that  in  the  vertical  11-12 ;  but.  when  th«  coun- 
ters are  not  in  action,  the  stress  in  11-12  is  always  equal  to 
the  load  at  12. 

Max.  positive  shear  in  10-12 

=  4  X  0.5  +  ^(1  +  2  +  •••  +  5)  =  15.5  tons. 

Full  panel  load  =  4  -f-  9  =  13  tons. 

.-,  Max.  stress  in  11-12  =  15.5  tons. 


W«WPtfi'Z"-^*'V>^- 


'-■■—-  .'->-3l*e?'59^fiaWWWSK».^^^;/;!W^-t.' 


( 


! 


78 


HOOFS  A\n  miiixsKs. 


Tlic  miiiiiiuiin  strras  in  any  vertical  is  e(|iial  to  tlic  maxi- 
mum iK'gativf  shear  in  the  stiction  cutting  it  and  two  chord 
luonihers,  or  it  is  equal  to  the  maximum  negative  shear  in 
the  panel  toward  the  cahutment  from  the  vertical,  or  to  a 
dead  jHinel  load,  whichever  is  the  greater.     Thus, 

Min.  stress  in  3-4    =  4  x  4,r»  -  0  =  18.0  tons. 
Mill,  stress  in  5-G    =  4  x  3.5  —  ^^  x  1  =  13.1  tons. 
Min.  stress  in  7-8    =  4  x  2.5  -  ^s  (1  +  2)  =  7.3  tons. 
Min.  stress  in  U-10  =  4  x  1.5  -  /^(l  +  2  +  3)  =  0.6  tons. 

IVit  the  stress  in  a  vertical  of  this  truss  cannot  be  less 
than  the  weight  of  a  dead  panel  load;  therefore 

Min.  stress  in  0-10  —  4  tons. 
Min.  stress  in  11-12  =  4  tons. 

We  may  now  collect  these  web  stresses  in  a  table  as 
follows ; 

Wkh  Stresses. 


Mkmiikrh. 


2-3 


Max,  Htrc8,ses 
Min,  8truiM(!H 


-82.48 
-25.38 


4-5 


-05.42 
-18.47 


fi-7 

-49.03 
-10.29 


>  9 


-35.10 
-  0.85 


10-11 


-21.85 
0.00 


9-li 

-U.87 
0.00 


VERTiCALS. 


Mkmiikm. 

8-1 

;,-'•, 

T  '* 

9  111 

11  [i 

Max.  stresses  .  .  ,  , 
Min.  stresses    .... 

68.5 
18.0 

40.4 

I'M 

36.2 

7..T 

24.0 
!.0 

15.6 
4.0 

-''Si-.'.^'t       -^J^^T^  ■ 


i^*-     "' 


I  to  tlic  maxi- 
11(1  two  chord 
tive  shear  in 
'tical,  or  to  a 
us, 

5. 

1  tons. 

7.',i  tons. 

J)  =  0.0  tons. 

iniiot  b(!  less 
re 


n  a  tal>le  as 


10-11 

9-1/ 

-^  11.87 
0.00 

-21.85 
0.00 

1  1(1 

II  I'i 

u.n 

10 

16.5 
4.0 

nitlDOE  TliUStWS. 


i!» 


Til  building  Howe  trus.ses  of  tiiubor,  it  is  usually  the 
practice  to  juit  in  all  the  dotted  diagonals  to  keep  the  tru.ss 
rigid.     This  is  not  done  in  metallic  trusses. 

I'rob.  60.  A  through  Howe  truss  has  11  panels,  each  10 
feet  long  and  10  feet  deep;  the  dead  load  is  SOO  Ihs.,  and 
the  live  load  is  1(500  1!,.h.  per  foot  per  truss:  tind  the  iiiaxi- 
iiiuni  and  minimuin  stresses  in  all  the  members. 

Ans.  Max.  stresses  in  lower  chord 

=  GO,  108,  144,  1G8,  ISO,  and  180  tons. 
Mill,  stresses  in  lower  chord 

-=  20,  3(;,  48,  5(5,  GO,  and  00  tons. 
Max.  stresses  in  main  braces 

=  _84.G0,  -08.71,   -53.83,    -40.0,    -27.18, 
—  15.37  tons. 
Mill,  .stresses  in  main  braces 

=  -28.28,  -21.5.",,  -13.85,  -5.13,  -0.00,  -0.00  tons. 

Max.  stresses  in  counters  =  —  4.31,  —  15.38  tons. 

Max.  stresses  in  verticals 

=  +  00.0,  +48.73,  +38.18,  +28.30,  +19.28  tons. 

Mill,  stresses  iu  vcvtioala 
=  -I-  yO.O,   1  15  '.'7,  +  9.82,  +  4.00,  +  4.00  tons. 

I'rnh.  61.  A  deck  Howe  trimN  liiis  12  [lanels,  each  10  feet 
long  and  10  feet  deep;  the  dead  load  is  000  lbs.,  and  the 
live  load  is  1200  llis.  per  foot  per  truss:  find  the  maxi- 
mum and  minimum  stresses  in  all  the  members. 

Ans.  Wax.  stresses  in  lower  chord 

=  49.5,  90.0,  121.5,  144.0,  157.5,  102  tons. 

Max.  stresses  in  main  braces 

=  -09.79,  -57.81,  -40.53,  -36.95,  -20.08,  -lO.Otons. 


9'<m-*i.T.mttmiiStiksSimS''ti^i*MeiKmi^-v  - 


MHH 


■MM 


HO 


ROOFS  AND   liliimsES. 


I    i 


Mill,  stresses  in  iiuiiii  bnuuts 

=  -  23.25,  -  l8.;i.S,  -  12.(;i),  -  G.ar.,  ().(),  0.0  tons. 

Max.  stresseB  in  counters  =  —  0.71,  —  b.-lO  tons. 
Max.  stresses  in  verticals 

=  1.5,  41.0,  33.0,  25.5,  18.5,  12.0,  7.5  tons. 

The  maximum  stress  in  any  vertical  of  the  deck  Howe 
truss  except  the  miildle  one,  is  equal  to  the  maximum  jmsi 
tivc  shear  in  the  section  cutting  it  and  two  chord  members, 
or  it  is  equal  to  the  maximum  jjositive  shear  in  the  panel 
away  from  the  abutment  from  the  vertical.  The  stresses  in 
the  verticals  need  to  be  very  carefully  tested. 

Pmb.  62.  A  deck  Howe  truss  has  10  panels,  each  14  feet 
long  and  14  feet  deep;  the  dead  load  is  714  lbs.  and  the 
live  load  is  2000  lbs.  per  foot  per  truss :  find  the  maximum 
and  minimum  stresses  in  all  the  ntcmbers. 

Ans.  Max.  stresses  in  lower  chord 

=  85.5,  152.0,  199.5,  228.0,  237.5  tons. 
Max.  in  main  braces 

=  -  120.55,  -  95.74,  -  72.89,  -  52.02,  -  33.37  tons. 
Min.  in  main  braces 

=  -31.73,  -  22.70,  - 11.70,  0.0,  0.0  tons. 
Max.  in  verticals 

=  9.5,  C7.9,  51.7,  3G.9,  23.5,  14.0  tons. 

Max.  in  counters  =  —  1.27,  —  16.22  tons. 

Proh.  63.  A  through  Howe  truss  has  8  panels,  each  18 
feet  long  and  24  feet  deep;  the  dead  load  is  806  lbs.,  and 
the  live  load  is  1500  lbs.  per  foot  per  truss:  fii.d  the 
maximum  and  minimum  stresses  in  all  the  members. 


"&  '■ 


UUinUK  TliVSSKS. 


HI 


0,0  tons, 
tuns. 


he  (lock  Ilowp 
iiaxiinum  posi 
liord  mciiibiTH, 
r  in  the  panel 
The  stresses  in 


Is,  eiieh  14  feet 
4  lbs.  and  the 
.  tlie  maximum 


33.37  tons. 


)anels,  each  18 
i  80G  lbs.,  and 
uss:    fii,4  the 
members. 


Art.  26,  Th«;  Pratt  Truss.— All  parts  <if  the  J'riut 
truss  arc  <>t  iron  or  steel,  the  verticals  bciiiL;  ennijircHsion 
members,  except  the  hip  verticals,  and  the  diiij^onals  ten- 
sion nicnibers.  The  niaxinmni  and  niininuiin  stresses  of 
an}'  member  jniiy  be  (k'tcrnuned  directly  from  a  single  Cipui- 
tion,  as  in  the  case  of  the  Howe  trnss  (Art.  2o). 

I'i-dIi.  64.  A  tliroii^di  Pratt  trnss,  as  a  through  railroad 
bridge,  Fig.  33,  has  10  panels,  each  14  toet  long  and  14  feet 


10         It        10'        8' 
Vlit.  33 


deep;  the  dead  load  from  formula  (1),  Art.  15,  the  live  load 
is  1800  lbs.  per  foot  per  truss :  find  the  maximum  and 
minimum  Htressos    :i  all  the  members. 

From  formula  (I;,  Art.  15,  the  dead  panel  load  per  truss 

=  (0x140  +  520)14  ^  12,460  lbs.  -=  G.23  tons  =  say,  6  tons. 

Live  panel  load  per  truss  =  16.2  tons  =  say,  16  tons. 

tan  0  =  1,  sec  6*  =  1.414. 

The  max.  ar.d  min.  chord  stresses  may  be  written  out  at 
once  (Arts.  10  and  'il).     They  are  the  following: 

Max.  stresses  in  lower  chord 

=  !)<>.0,  90.0,  176.0,  231.0,  264.0  tons. 

Min.  stresses  in  lower  chord 

=  27.0,  27.0,  48.0,  63.0,  72.0  tons. 


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82 


HOOFS   A.\h   lililbGES. 


Wi;ii  Stukssks. 


Tlic  wfl)  stro.sscs  may  he  fuiiiul  exactly  as  in  Prob.  60. 
Th(!  I'olluwiiig  arc  (lie  o<iiiatiui!s  for  a  few  of  tlie  iiu'ml)ers: 

Max.  stress  in  2-3  =  -  [4.0  x  22]  1.41  =  -  139.5  tons. 
Max.  stress  in  3-0 

■■=  \}\.'>  X  ()  -f-  l.O^l  +  2  +  ■..  +.  S)]1.41  =  110.8  to'is. 
Max.  stress  in  0-8 

=  [2.5  X  6  +  l.G(l  +2  +  ...  +-  7)11.41  =~.  84.3  tons 
Max.  stress  in  7-8 

-:  -[i.5  X  0  +  1.0(1  +.  2  +  ...  +-  ())]  =  -42.0  tons. 
Max.  stress  in  11-12 

-  -  [-  0.5  X  0  +  1.0(1  +.  2  +  ...  +  4)]=  -  13.0  tons. 
Min.  stress  in  5-8 

=  [2^  X  0  -  1.6(1  +  2)]1.41  =  14.4  tons. 
Max.  stress  in  10-11 

=  [-  i  X  0  +  1.0(1  +  2  +  ...  4-  4)]1.41  =  18.5  tons, 

Thus  we  find  the  stresses  in  the  following  table : 


Weii 

Stuesses. 

\!'.mp;krm. 

•:-'! 

S-fi 

1 

5-S            1-10 

9-12 

S  9 

10-11 

Max.  .strtsses 
Min.  stresses 

-l:!9.'. 
-  .-Js.l 

+  110.8 
+  27.4 

+81.3  1  +(10.0 
+  14.4        0.0 

-f  .S8.0 
0.0 

+  0.8 
00 

+  18.6 
0.0 

Max.  stresses  in  Ihe  vertieals  =+-22.0,   -50.8,   -42.0, 
-  27.0,  -  iii.b  tons. 


HRWUE  TRUSSES. 


m 


1  ill  I'loi).  (;o. 

Ik!  iiu'iiihers: 


Proh.  65.  A  tliio.ij,'].  I'liitt  truss  lias  H  i.anels,  ea'-I.  10 
fffi  Inn-  and  10  feet  (l.-cp;  tilt;  dead  load  is  IL'OO  Iks.,  and 
tlh'  iiv..  lead  is  L'OOO  lbs.  per  fout  per  truus :  find  the 
stresses  in  all  the  'neiuhers. 

Ans.  Max.  sticsses  in  lower  eliord 

=  S0,  SO,  Ml,  liH,',  L'L'-l,lM(M,oi.s. 
Max.  in  main  diagonals 

=  -ll;U,  -i-OI..'-,,  +7l.r,,  +50,8,  4..'W.4,  +19.2 tons. 
Min.  in  main  diaj^onals 

=  -  IL'.;},  -I- 1^2.{^,  +  2\j>,  +  9.2, 0.0,  0.0  tons. 

Max.  in  eounters  ^  -f-  4.4,  +  J9.L'l  tons. 
Max.  in  verticals 

=  +  IG.O,  -  r>0.7,  -  37.4,  -  LT).],  -  13.6  tons. 

/Vo/>.  66.    A   deck  Pratt  icuss,    Fij,'.  34.  has  10  panels, 
each  lli  feet  long  and  12  feet  deei>;  the  dead  load  is  833 


18.5  tons, 
ible: 


S  9 

lii-tl 

+0.8 
0.0 

+  18.5 
0.0 

59.8,   -42.0, 


ll>s.,  and  the  live  load  is  1667  lbs.  per  foot  per  truss:  find 
the  stresses  in  all  the  members. 

Ans.  Afax.  stresses  in  lower  chords 

=  0.0,  67.5,  120.0,  157.5,  180.0  tons. 
Max.  in  main  ties  =  95.1,  75.4,  57.1,  40.2,  24.6  tons. 
Min.  in  main  ties  -  31.7,  23.2,  13.4,  2.1,  0.0  tons. 
Max.  in  eonnters  =  10.6  tons. 
Max.  in  verticals 

=  -  75.0,  -  67.5,  -  53.5,  -  40.5,  -  28.5,  -  20.0  tons. 


■  k 

I 


f 


( 


84 


HOOFS  A  XI)   lilllDGKs. 


I'fth.  67.  A  ik'ck  I'latt  truss  liiis  i»  panels,  caoh  IS  feet 
long  mid  L'l  Icct  dccj);  tlio  dead  load  is  .'.(MMlis.,  and  flu- 
live  load  is  !()(»»  His.  per  foot  per  truss:  Hud  the  stresses 
in  all  the  nienibcrs. 

Ans.  Max.  stresses  iu  lower  chords 

=  0.0,  JO.-),  70!).<)1,  1)1.1,  lOl.;}  tons. 
Ma.\.  stresses  iii  main  lies 

=  (•.:.:.,  .-.1.0,  -.Mli,  L'4. 1,  12..-.  tons. 
Mill,  stresses  in  main  ties 

^  L'L'..-.,  l.-..r.,  7..-.,  0.(»,  0.0  tons. 
Max.  stresses  in  counters  —  1.0,  lL'..j  tons. 
Mi'.x.  stresses  in  verticals 

=  _  no.s,  -  54.0,  -  41..-.,  -  .30.0,  -  10.5  tons. 

The  chord  stresses  are  the  same  in  the  Howe  trusses,  and 
also  in  the  Tratt  trusses,  whether  the  load  be  jilaced  on  the 
top  or  the  bottom  chord. 

Art.  27.  The  Warren  Truss  with  Vertical  Sus- 
penders.—  The  left  half  of  this  truss  is  represented  in 
Fig.  35,  in  which  the  web  bracing  consists  of  equilateral  or 


isosceles  triangles  and  vertical  ties  or  suspenders.  In  a 
through  truss,  each  of  the  verticals  .3,  5,  7,  9.  simply  carries 
a  paiud  load  from  the  lower  chord  to  the  upjier,  while  the 
verticals  I,  (>,  8,  10,  are  only  to  siipjiort  and  stiffen  the 
upper  chord.  If  the  truss  is  a  deck,  the  last  named  ver- 
ticals become  posts. 


■  ''slSSl*^'" 


BlilDGE  TliUSSEfi. 


85 


caoh   is  feet 

11)8.,  anil  tlir- 

the  stresses 


-  lO./i  tons. 

?  trusses,  ami 
[)lacetl  on  the 


irtical  Sus- 

'l)resent('(l  in 
3quilateral  or 


nders.  In  a 
imply  carries 
er,  while  tlie 
I  stiffen  the 
t  named  ver- 


Prnh.  68.  A  through  Warren  tnbTi  with  vertical  ties, 
one  half  of  which  is  shown  in  Fig.  .};■>,  has  16  panels,  each 
S  feet  long,  its  braces  all  forming  equilatenil  triangles;  the 
dead  load  is  given  by  formula  (2)  of  Art.  15,  the  live  load 
is  loCM)  lbs.  per  foot  per  truss:  find  the  maximum  and 
minimum  stresses  in  all  the  members. 

The  dead  panel  load  per  truss 

_/7x  128  +  600\y     ^oeiiu        •>  * 
=■' —     -  18=5984  lbs.  =  3  tons. 

Live  panel  load  per  truss  =  — - ^-?  =  G  tons. 

tan  e  =  .577 ;  sec  d  =  1.1547. 

The  chord  stresses  may  be  written  out  at  once  by  chord 
incremc  rts. 

For  the  verticals  the  maximum  r.tress  is  evidently  the 
dead  and  live  panel  load,  or  9  tons;  the  minimum  stress 
is  the  dead  panel  load,  or  :i  tons. 

The  following  are  the  equations  for  iletermining  the 
stresses  in  a  few  of  the  members: 

Max.  stress  in  2-t  =  7.50  x  9  x  .677  =  38.9  tons. 
Max.  stres3  in  4-6 

=  [7.50  +  G.50  +  5.50]9  x  .577  =  101..3  tons. 
Max.  stress  in  3-5 

-=  -[7.50  +  (5.50]  X  9  X  .577=  -  72.7  tons. 
Max.  stress  in  2-3  =  -[7.50  x  9  x  1.154]  =  -  77.9  ton.s. 
Max.  stress  in  ,5-() 

=  [4.5  X  3  +  /,  (1  +  2  4-  3  +  ...  4- 12)]  1.154 

=  42.75  X  1.154  =  +  49.3  tons. 
Min.  stress  in  7-8 

=  [2.5  X  3  -  T-\  (J  +  2  +  3  +  -  +  5)]  1.154  =  +  2.2  tons. 
lu  this  way  all  the  stresses  may  be  found. 


m 


ft 
I 


— ^ 


86 


HOOFS  A\h  llllIDaEs. 


riiiiiiii 

SriiKKSKs. 

Memukim. 

«-6 

.V7 

I  9 

9-9 

2-4 

4-8 

6-8 

S-H) 

Max.  strpssi's 
Mln.  Htrir.HM'.H 

-72.7 
-24.2 

i24.r. 
-  41.;. 

~  IW.*. 
-  51.9 

-  lfifi.2 

-   .V..4 

+  3^.9 
+  l.'-.O 

f  101.3 
+   8S.S 

+  142.  S 

+   47.6 

+  111;.'- 

Wkii  Stbksses. 


Mrmiiikk. 


Max.  rtr«»e!i . 
Mln.  ttrttaM<8  , 


2  8 

-77.9 
-2110 


3-4 

+  117.9 
■^  22. 1 


-^>*.4 
-17.7 


+  49.3 
+  13.0 


C-7 

-40.6 

-  7.S 


.S-9 


-(-•32.6     -24.7 
+    2.2  I  -I-   3.9 


-»■  17.3 

10.1 


We  see  fn)iii  tlii.s  tiiblo  of  wel)  stre.sses  tliat  tlie  niciiibors 
8-9  and  9-10  should  lie  cai)al)lo  of  resisting  both  tension  and 
eoiiipression,  and  hence  siiould  be  eouiiter  braces  (A,  .  1^; 
and,  of  course,  tlie  same  is  true  for  the  right  half  of  the  truss. 

J'rob.  69.  A  through  Warren  truss  with  vertical  ties,  like 
Fig.  ;jr>,  has  20  panels,  each  10  fe.'t  long,  its  braces  all 
forming  equilateral  triangles;  the  dead  load  is  given  by 
formula  (li)  of  Art.  lo,  the  live  load  is  lOCK)  lbs.  per  foot 
l)er  truss :  find  the  niaximum  and  minimum  stresses  in  all 
the  membeis. 

Aus.  Max.  stresses  in  upper  chord 

=  -i;{r>.o,  -2-10.0,  -3ir,.o,  -sco.o,  -375.0  tons. 

Max.  stresses  in  lower  chord 

=  +  71.;i,  +191.3,  -f- 281.3,  +341.3,  +371.3  tons. 
Max.  web  stresses 

=  -142.;-.,   +128.0,   -113.9,   +100.2,  -87.1,  +74.4, 
-  (52.2,  +  50.4,  ~  39.1,  +  28.3  tons. 
Min.  web  stresses 

=  -54.8,  +48.G,  -41.9,  +34.7,  -27.1,  +19.0,  -10.5, 
+ 1.5,  +  8.0,  -  17.9  tons. 


m 


6-8 

8-11) 

+  1«.S 
+   47.6 

+  Kill.'- 
+    M.- 

S 

S(-9 

'    +           -- 

.6 
2 

-24.7 
+  8.8 

the  nicnibors 
li  tension  and 
ices  (A,  .1^; 
f  of  the  truss. 

tieal  ties,  like 
ts  braces  all 
is  given  by 
11)8.  per  foot 
tresses  in  all 


375.0  tons. 
?71.3  tons. 
■87.1,  -1-74.4, 

-19.0,  -10.5, 


HHlltaK  TRUSSES. 


87 


Prnh.  70.  A  tliroii;j;li  Warren  triiss  with  vertical  ties, 
lik<'  V'y^.  ."r>.  has  10  panels,  each  12  feet  long  and  VJ 
feet  tlcep,  its  braces  all  forn.ing  is().s<'eles  triangles;  th(^ 
dead  and  live  loads  are  100((  lbs.,  and  I'OOO  lbs.  per  loot  per 
trn.ss:  tind  the  niaxiniuni  and  niinininni  .stresses  in  all  the 
meiubers. 

Art.  28.  The  Double  Wairen  Truss,  or  Double  Tri- 
aitfjiilar  Truss,  or  Ginlcr.  -  This  truss,  shown  in  Fig.  ofi, 
has  two  systems  of  triangular  bracing,  the  one  system 
represented  in  full  lines  and  tlie  other  in  dotted  Hues,  the 


J s     r,     7     ."    ti ?'    7'    s'    .1'    r 


chords  being  common  to  both  systems.     The  roadway  may 
be  carried  either  npon  tho  upper  or  the  lower  chord. 

In  all  the  trusses  hitlierto  considered,  a  vertical  section 
taken  at  any  point  of  the  truss  will  cut  only  three  members ; 
but  in  the  double  Warren  truss  this  is  not  the  case,  since  a 
vertical  section  will  in  general  cut  fimr  mend)ers,  causing 
the  problem  at  first  to  seem  to  be  indeterminate  (Art.  7). 
This  is  obviated,  however,  by  the  fourth  condition  that: 
the  truss  is  ecpiivaleut  to  the  two  trus.ses,  shown  in  full  and 
in  dotted  lines,  welded  into  one.  Thus,  the  loads  at  fi,  10, 
10',  G'  are  carried  to  the  abutments  by  the  diagonals  drawn 
in  full,  while  the  loads  at  4,  8,  12,  8',  4'  are  carried  by  the 
dotted  diagonals.  The  chord  and  web  stresses  are  retulily 
found  as  in  a  simple  triangular  truss,  assuming  each  system 
as  independent.  This  truss  is  used  both  as  a  riveted,  and 
as  a  pin-eonuccted  bridge. 


.-^"uriAiMr^^ii 


.jMMH 


-^TJTflT 


•f 

I 


} 


i 


aaMH 


88 


nnoFs  AND  unnxms. 


Pmh.  71.  A  (louhlo  Wan-on  trus.s,  Fig.  m,  as  a  dock  rail- 
road hiid^'o,  li.xs  Id  paiiols,  oach  14  foot  lonj,'  and  14  foot 
dooii;  tlic  d,.;i(l  load  is  Lfivcii  l.y  foniiida  (L')  of  Art.  ir»,  the 
livo  load  is  L'lMIO  ilw.  per  loot  per  truss:  tind  the  stresses  in 
all  the  members. 

The  dead  panel  load  per  truss 


V 2 )      "^  l^OGO  lbs.  =  5.5  tons. 


Live  panel  load  i)er  truss  =  14  tons. 

tan  5  =  1,  sec  5  .=  1.414. 
The  followiiif,'  aro  the  etpiations   for  determining  the 
stros.sos  in  a  few  of  the  niend)ers: 

Max.  stress  in  l-H  =  -  2  x  1!).;")  x  1  =  -  39  tons. 
Max.  stress  in  .3-5  =  -  [2.0  +  2.5  4- 1.5]  19.5  =  -  117  tons. 
Max.  stress  in  (Mi  =  [2.5  +  4.0  -|-  3.0]  19.5  =  + 185.2  tons. 
Max.  stress  in  .'M! 

^  [1.5  X  5.5  +-  1.4  (1  +  8  +5  +-  7)]  1.414=  +  43.3  tons. 
Min.  stress  in  5-.S  ^  [5.5  -  1.4  x  2]  1.414  =  +  3.8  tons. 
Thus  the  following  stresses  are  determined : 

Upper  Chord  Stresses. 


Mkmheiw. 

13 

3-5 

.^•7 

7-9 

9-1! 

Max.  stroRses    .  .  . 
Min.  stresses    .  .  . 

-3i».0     -117.0 
-11.0     -  .3.S.0 

-175.5 
-  49.5 

-214.5 
-  00.5 

-234.0 
-  00.0 

Lower 

niioRi)  St 

KESSRS. 

.Mkuiiers. 

2-1 

4-r. 

fi-S 

s-in 

1(1-12 

Max.  8lre.s.ses   .  .  . 
Mill,  stri'sscs    .  .  . 

+48.8 
+  1.3.8 

+  120.8 
+  35.8 

+  18.5.2 
+   52.3 

+  224.3 

1-  03.3 

+24.3.8 

+   08.  S 

a  a  (lock  rail- 

and  14  foot 

:'  Alt.  IT),  the 

lio  stresses  in 


5  tons. 


nnining  the 

ma. 

=  - 117  tons. 

-f- 185.2  tons. 

-  43.3  tons. 

-  3.8  tons. 


9 

9-11 

4.5 
0.5 

-234.0 
-  00.0 

n 

1(1-12 

1.3 

5.3 

+243.8 
+  08.8 

BlUDdK  riiUSSES. 


89 


Wkh  Strksshs  (Dottki) 

"ivSTEM). 

MKMItKltf*. 

i-i 

1  .•. 

.'i  H 

s-K 

!t-|-.' 

.Max.  stresses 

.Mill.  .Htreases 

+  55.2 
+  15.0 

-55.2 
- 15.0 

+  31  6 

+   3.8 

-31.5 

-  ;i.8 

+  n.» 

-ll.O 

Wkii  Strksskh 

[FiM,  System). 

.Mkmiikk". 

■i-:\ 

:!-■! 

<;-T 

7-10 

10-11 

Max.  stresses 

-00.0 
-10.4 

+  43.3 

+  0.7 

-43.3 
-  0.7 

+  21.7 
-  4.1 

-21.7 
+  4.1 

Mill,  stresses 

Prob.  72.  A  double  deck  Warron  truss,  like  Fig.  3f),  has 
12  panels,  each  10  feet  lon<,'  and  10  foet  deep;  the  dead  load 
is  SOO  lbs.,  and  the  live  load  is  1(500  lbs,  {ler  foot  per  truss : 
tiiid  the  stresses  in  all  the  nieinbers. 

Aus.  Max.  stresses  in  upper  chord 

=  -  30,  -  90,  -  138,  - 174,  -  198,  -  210  tons. 

Max,  stresses  in  lower  chord 

=  30,  9(),  144,  180,  204,  21G  tons. 

Max.  stresses  in  dotted  diagonals 

=  +  42.4,  -42.4,  -j-27.3,  -27.3,  -f  14.1,  -14.1  tons. 
Max.  stresses  in  full  diagonals 

=  -60.9,  +34.9,  -'M.9,  +20.7,  -20.7,  +8.5  tons. 
Min.  stresses  in  dotted  diagonals 

=  + 14.1,  - 14.1,  +  G.G,  -  6.6,  -  2.9,  +  2.9  tons. 

Min.  stresses  in  full  diagonals 

=  -17.0,  +10.4.  -10.4,  +1.7,  -1.7,  -8.5  tons. 


I 


90 


HOOFS  AMt   ItllllX/KS. 


Pnth.  73.  A  (lowltlc  tliruii^'li  Wjiin-ii  Iniss,  like  Fij».  .'{f., 
IliVS  1(1  |i,llirls,  iMcIl  10  IVct,  li.lij,'  iiikI  10  r,(.|,  .ItTl-;  tlic  .Icml 
and  liv."  loiids  arc  SOO  ll.s.,  and  L'OCM*  \h».  per  foot  per  Uiisa: 
find  tho  stresses  in  all  tlie  inendx-rs. 

Ana.   Max.  stresses  in  npiier  chord 

=  -  ;jr>.  -  i>i,  -  i;;.!,  -  u\\,  -  ito  tons. 

Max.  stresses  in  lower  chord 

=  +  L'S,  +  St,    f  IL'C,  +  loj,  -»-  ICS  tons. 
Max.  stresses  in  dia.i,'onal  ties 

=  +  -l!)..|,   +■■!'.).(•.,   +.'{1.1,  4-lil.'.r,.  +ir>.r.,  +H4ton8. 
Min.    stresses  in  dia<,'onal  ties 

=  +  14.1,  +11.;!,  +7.1,  +L'.8,  -IM),  -8.4  tons. 
J'rnh.  74.  A  donlile  thr.)ni,'h  Warren  trn.ss,  like  Fi;L'.  .'!(), 
has  IL'  panels,  each  1".  feet  Ion;,'  and  1".  feet  deep;  tlie  dciul 
load  is  given  hy  fornmla  (.'i)  of  Art.  15,  the  livi^  load  is 
L'OOO  lbs.  per  foot  per  truss  :  find  the  stresses  in  all  the 
mend)er8. 

Art.  29.  Tho  Whipple  Truss.  -This  truss,  shown  in 
Fig.  ;{7,  consists  of  two  simple   Tratt  trusses  combined. 


;5*    10' 
KiB.  37 

The  ties  in  the  web  system  extend  over  two  panels,  and  it 
is  therefore  often  called  a  "  double  intersection  Pratt  truss."* 
The  advantage  over  the  I'ratt  for  long  spans  is  that  it  has 
short  panels,  wliile  keeping  the  inclination  of  the  diagonals* 
at  about  45°. 

•  Called  also  the  Linville  Truss. 


mam 


till'  ilciid 

[M'l  truss: 


uuiixu:  TurssKs. 


91 


Tliis  tnisH  WHS  at  one  tinu-  iiKtre  coiiiiiioii  lliaii  any  otiicr 
III  Aiiicricaii  liriih^c  |ii;iili(  c.  I),  is  slill  iiscil  Ili(iii;,'li  vi-ry 
ran-ly  tor  lii^,'liwii_v  hiidgrs;  lor  railway  luiilKfs  it.  is  as  a 
rule  ivvoidnl  by  tin-  licst  prai'tico. 

It  is  assiiiiu'il  that  tliis  truss  is  oiiuivalciit  to  tli«>  two 
trusses  shown  in  full  and  in  dotted  linos,  and  that  each 
system  aets  indei)endeiitly  and  carries  only  its  own  loads  to 
the  abiitinents;  but  in  the  case  of  the  web  nienibers,  this  is 
not  stn'rth/  true.  If,  however,  the  truss  is  built  witli  an  even 
miiiiber  of  panels,  the  error  in  the  web  stresses  (  btained  on 
the  assumption  of  independent  systems  is  prolialily  very 
small.  With  the  chord  stresses  there  is  no  anibij,'Mity  ;  the 
idiord  stresses  are  a  maximum  for  a  full  load,  and  may  bo 
found  as  usual  by  chord  increments,  or  by  monu'iits.  The 
vertical  shear  will  be  tlie  same  whether  the  load  is  applied 
at  the  top  or  at  the  bottom  chord. 

Pn>h.  75.  A  througli  Whipple  truss,  Fig.  J^",  has  12  j)anel8, 
each  10  feet  long  and  I'O  feet  deep;  the  dead  load  is  ToO  lbs., 
and  th(>  live  load  is  12(M>  lbs.  per  foot  per  truss:  find  the 
stresses  in  all  the  meml)er.s. 

The  dead  panel  load  per  truss  =  S.T.T  tons. 

The  live  panel  load  per  truss   =  CO  tons. 

tane  =  l:  seed  =  1.414:  tan  0'  =  0.5:  secfl'  =  1.118. 

The  following  are  the  cipiations  for  finding  tlie  stresses  in 
a  few  of  tlie  members : 

Max.  stress  in  4-G  =  3  x  9.75  x  0.5  =  14.(»  tons. 
Max.  stress  in  0-8  =  14.G  +  2}^x  9.75  x  1  =  39.0  tons 
=  —  stress  in  1-3. 

Max.  stress  in  9-11 

=  -  [39.0  +  (2  +  1^1  +  i)9-75]=  -  87.8  tons 
=  stress  in  11-13, 


•-'i 


r 

i 

1 


92 


ittutFs  .i.v/>  itni'xiKs. 


sinro  for  a  iinifdnn  Icnul  the  (li.i^fonals  niocMiii},'  lln'  ii|i|«M' 
olionl  Im'I  wtfll  ',)  iillil  '.>'  ;ili'  lint   ill  iirtidil. 

Mux.  Htrt'HS  ill  l-l  — .'{  X  'Xlii  X  l.llK  j^i'W."  tons. 
Mftx.  Htri'sfl  in  MJ 

=  [l'.J  X  ;J.7r.  +  ,«^  (L'  -f.  4  +  0  +  8  -j  10)]  X  1.414 

=  lil.;t7r.  X  l.JM-^.'M.ntons. 
Mux.  stress  in  1  -L'  =  -  •.i'J.'J't  -  LM.STfi  =  -  53.(5  tons. 

Thus  tho  following  stresses  are  (letorniiiied . 
{'iii>ni>  Stiikshes. 


Mkmiikkk. 

Max.       . 
.Mill.       . 

•:-i 

4  a 

ll.ll 

r..ti 

0* 

»-IO 

ii;-i!j 

7:i.l 
2H.I 

IMJ 

ll-l!l 

H7.8 
:).'{.8 

ll.tM) 
II.IMI 

:M).o 

15.0 

58.5 
22.6 

«2.0 

;ii.8 

Stukhhi:*  i; 

1  niA<i 

INAI.H. 

VKMIirilH. 

1. 

1  II 

:ls 

,'l  111 

7-1 'J 

«  U 

11-IJ' 

IH   111' 

Mnx.      . 
Mill.       . 

n2.7 

12.*( 

;u..-i 
i:i.;i 

28.  .T 
0.0 

22.1 
0.5 

ICO 

2.5 

11.1 
0.0 

0.4 

0.0 

1.0 
0.0 

Stiikssks  IN   Posts. 


.MlUBEItK, 

1-2 

!•-» 

&-« 

7-S 

11-10 

11-12 

IS- 14 

Max.     . 

Mill.      . 

53.0 
20.0 

20.0 
7.0 

15.0 
4.0 

11.8 
1.8 

7.0 
0.0 

4.6 
0.0 

1.1 

0.0 

Pi'ob.  76.  A  tlirough  Whipple  truss,  Fig.  38,  has  16 
panels,  each  10  feet  long  and  L'O  feet  deep ;  the  dead  and 
live  loads  are  800  lbs.,  and  IGOO  lb;4.  per  foot  per  truss; 
find  the  stresses  in  all  the  members. 


liUWUE  TIlUSHEli. 


98 


I ; 


/:      tn' 


0 

H7.H 

8 

.w.a 

IJ' 

18  III' 

.4 

1.0 

.0 

0.0 

in  14 

1.1 

0.0 

Ans. 


to     u     i;      la     la     lo 
Kilt,  as 

Ciiiiiih  STnuHsiiA. 


Mkhhkm. 

4  11 

21. 

n  > 

■-1(1 

Ii)-l-i 

111-14 

M-ll> 

IB-IS 

1»  IT 

Max.       . 
Mill.  .     . 

(HI. 

102. 

i:J2. 

44. 

IM. 

r.2. 

174. 

OH. 

IHtl. 
02. 

1',I2. 
«14. 

HrnKssKS  iv  Fri.i,  DiAfioxAi.s. 


Mknhkbh. 

Ml 

.'.-in 

n-u 

l:t-ls 

n-14' 

Max.     .     .    . 
Mill.     .     .     . 

r.0.4 

10.8 

4;i.8 

12.7 

20.7 
4.2 

17.0 
0.0 

f).7 
0.0 

STiu:asK8  IN  DoTTKit  Diauonals. 


MRMHKB!). 

i-« 

:t  ■< 

7  12 

11-16 

15-1«' 

15-12 

Max.    .     . 
Mill.     .     . 

r.:t.7 
22.0 

51.0 
10..-! 

30.8 

8.4 

2.1.3 
0.0 

11.3 
0.0 

0.7 
0.0 

Stukssks  in  I'osTH. 


Mkmkkkh. 

1-2 

!14 

.'. « 

7-» 

20.0 
0.0 

U-10 

11-12 

lit  14 

l.Vl« 

n-i» 

Max.      .     . 
Mill.       .     . 

1K).0 

;io.o 

30.  f) 
11.5 

;ii.o 

0.0 

21.0 
3.0 

10.5 
0.0 

12.0 
'  0.0 

8.0 
0.0 

4.0 
0.0 

r 

I 

< 


ri 


\ 


94 


HOOFS  Ash  nniunKs. 


Proh.  11.  A  (leek  Wliipplc  truss,  like  Fi^.  38,  has  Ifi 
panels,  each  1L'  feet  long  ami  L'l  feet  »l<'cji;  tiic  dead 
load  is  THO  ihs.  por  foot  per  truss,  and  tlif  liv  ■  load  is 
-(HH)  lbs.  per  foot  por  truss:  Hiid  the  stresses  in  all  the 
member^'. 

Art.  30.  The  Lattice  Truss,  or  Quadruple  Warren 
TrusiS.*  —  This  truss,  Fig.  31),  contains  four  web  sysloinb 


15     g     7     !)     fl    19   /5    n   10  «/ 


l»     Ci     8    iu   U   Ik  W  18  to  in   i»'  18'  10'  11,'  i,'  lu'  8'   0'    U'    ^  j' 

welded  into  one.     It  is  built  only  as  a  short-span  riveted 
structure. 

In  determining  the  .stresses  in  the  different  nit-iubers  of 
this  tiass,  an  ambiguity  arises  from  two  eanses :  (1)  the 
web  members  being  riveted  together  at  each  intersection, 
the  different  systems  cannot  act  independently ;  and  (2)  two 
of  the  systeinr,  are  not  symmetrically  placed  in  reference  to 
the  center  of  the  truss.  Thus,  if  etpial  loads  1m,  placed  at 
each  joint,  the  abutment  at  2  will  carry  more  than  half  of 
the  load  on  the  system  4;  12,  20,  16',  8',  and  less  than  Ijalf 
of  the  load  on  tli"  system  8, 13,  20',  12',  4'.  I?nt  it  simplifies 
the  solution  toassunu'  that  each  system  acts  independently, 
and  is  symmetrical  in  reference  to  the  center  of  the  truss. 
The  chord  and  web  stresses  are  then  readily  found,  as  in  a 
simple  triangular  truss. 

*  A'l  rtimble  sy  teiiis,  such  .is  this  aii<l  tlu"  Wliippip,  tiwiii};  lo  tlio  iiido- 
lorniiiiaU'  I'li.inioter  of  the  strains,  an-  imw  usually  avoided  liy  tlu;  best 
practice. 


m 


...Jifil 


BltlDGE  TRUSSES. 


95 


Prob.  78.  A  through  lattice  truss  contiiiuing  four  web 
systems,  as  shown  in  F'ig.  -'^9,  has  20  panels,  each  10  feet 
long,  the  depth  of  the  truss  is  20  feet;  t!ie  dead  load  is 
given  liy  formula  (2)  Art.  15,  the  live  load  is  2000  lbs.  per 
foot  per  truss :  find  the  stresses  in  all  the  members. 

Dead  panel  load  jier  truss  =  5  tons. 

Live  panel  load  per  truss  =  10  tons. 

tan  e~l,  see  e  =  1.414. 

Note.  —  In  fiiuliiif;  the  shear  for  a  web  ineinber  (hie  to  the  (le<ad 
load  or  to  tlie  live  hiiul  eoveriiig  the  whole  trii.s.s,  a  slight  difticulty 
arises,  owing  to  liie  two  ..nsyimnetrical  systems  above  referred  to. 
Tills  shear  may  be  found,  either  by  taking  the  algebraic  sum  of  the 
greatest  positive  and  the  greatest  negative  shears  (Art.  22),  or  by 
simple  inspection  of  the  trus.s. 

Thus,  the  dead  load  shear  in  o-lO  is  the  weight  of  the  two  panel 
loads  at  the  joints  10  and  18,  because  the  .system  to  which  r>-10  be- 
longs is  ayniinetncal  with  reference  t  the  center  of  the  truss ;  but 
the  shear  In  7-12  Ir  not  the  weight  of  the  two  panel  loads  at  the  iolnts 
12  and  20,  because  this  system  Is  not  synniietrlcal  in  reference  to  the 
center.  If  the  joint  20  were  at  the  center  22,  the  shear  for  7-12  woidd 
be  1 J  panel  loads  ;  but  If  It  were  at  the  joint  is,  llu'  shear  would  be 
2  p.int'l  loads.  Being  midway  between  the  joint.s  18  and  22,  the  shear 
for  7-12  is  the  mean  of  l.J  and  2  panel  loads,  or  1.75  panel  loads  ;  and 
so  for  the  shear  in  any  other  web  member. 

By  chord  increments,  we  have 

Max.  stress  in  2-4  =  2  x  15  x  1  =  30  tons. 

Max.  stress  in  4-0 

=  (2  +  1.75  +  2.75)  X  15  =  97.5  tons. 
Max.  stress  in  0-8 

=  97.5  +  (1.5  +  2.5)  X  15  =  157.5  tons. 
Max.  stress  in  l-*5  =  2.5  x  15  =  37.5  tons. 
■     Max.  stress  in  3-5  —  (2.5  +  2  x  2.25)  x  15  =  105.0  tons. 


■■  ii,»np^fig»HWWi— »» 


I 


i 


96 


ROOFS  AND  nuiDGKS. 


Maximum  Web  Stues.sk.s. 


Tho  maximum  stress  in  ;\-S  will  bo  when  the  dead  and 
live  loads  cover  the  whole  sy.st,eni  to  which  3-8  belongs. 
Likewise  for  the  menilmr  o-lO.  The  maximum  live  load 
stress  in  7-lL'  will  be  when  the  live  pane!  loads  are  at  the 
joints  IL',  'JO,  1(5',  8',  since  these  are  the  only  loads  which 
a.;t  in  the  .system  to  which  7-12  belongs,  on  the  right  of 
7-12.     For  the  dead  load  stress,  see  Note. 

Max.  stress  in  3-8 

=  2.25  X  15  X  1.414  =  47.7  tons  =  -  stress  in  .1-3. 
Max.  stiass  in  5-10 

=  2  X  15  X  1.414  =  42.4  tons  =  -  stress  in  2-5. 

Max.  stress  in  7-12 

=  [1.75x5+ .JO (3+7^11^.iri)-]l  414^3-.  g^^^g 

=  —  stress  in  4-7. 
Max.  stress  in  9-14 

=  [1.5  x  5  +  ^  (2  +  0  + 10  + 14)]  1.414  =  33.2  tons 
=  —  stress  in  (J-l). 

Thus  the  following  stresses  are  determined: 
Ui'PKit  Ciioni)  Stresses. 


Memhkrs. 

1-3 

S-A 

.V7 

T-9 

9-11 

Max.     .     .     . 
Min.     .     .     . 

37.5 
12.5 

106.0 
35.0 

106.0 
55.0 

217.5 
72.6 

202.6 

G7.5 

MiMIIKUS. 

Max.     .     .    . 
Min.     .     .    . 

11-13 

l.S-l,'> 

15  IT 

17  111 

19-21 

375.0 
125.0 

300.0 
100.0 

.330.0 
110.0 

352.5 
117.5 

.'!07.5 
122.5 

m^ 


!  dead  and 
S  l»elong,s. 
1  live  load 
are  at  the 
latls  which 
le  right  of 


a  ^-3. 


tons 


tons 


9-11 


202.5 


19-21 

375.0 
125.0 


U JUDGE  TRUSSES. 
Lower  Ciiorp  Stresses. 


97 


.Memiip.iis. 

2-t 

■)-■'. 

0-1 

^-m 

lU-13 

Max.     .     .     . 
Mill.      .     .     . 

30.0 
fl.O 

07.5 
;{2.5 

157.5 
52.5 

210.0 
70.0 

255.0 
85.0 

Mkmiikus. 

IJ-U 

11  111 

IC-li 

H-2() 

300.0 
120.0 

•M  22 

Max.     .     .     . 
Mia.      .     .     . 

202.5 
07.5 

322.5 
107.5 

345.0 
115.0 

307.5 
122.5 

Web  Strkh,se.s. 

Max.  stre.ss  in  A-4  =  58.4,  in  l-C  =  M.O,  in  3-8  =  47.7, 
in  5-10=42.5,  in  7~1L'=.'{7.8,  in  \)-U=3:i.2,  in  11-1G=l'8.7, 
in  K5-18  =  24.0,  in  15-20  =  20.2,  in  17-22  =  1G.3,  in  19-20' 
=  12.3,  in  21-18'  =  8.5,  in  19'-1G'  =  5.4  tons. 

Min.  stress  in  A-4  =  10.5,  in  1-6  =  17.7,  in  3-8  =  15.9, 
in  5-10  =  14.2,  in  7-12  =  11.8,  in  9-14  =  9.2,  in  11-16  =  6.8, 
ill  i;}-18  =  4.3,  in  lJ-20  =  1.3,  in  1.7-22  =  -  2.1,  in  19-20' 
=  -  5.3,  in  21-18'  =  -  8.5,  in  19'-16'  =  -  8.9  tons. 

Max.  stress  in  1-.>:1  =  37.5,  in  ^1-2  =  112.5  tons. 

Min.  stress  in  1-^1  -■  12.5,  in  ^1-2  =  37.5  tons. 

We  see  from  these  results  that  the  10  diagonals  14-17 
and  17-22, 16-19  and  19-20',  18-21  and  21-18',  20-19'  and 
19'-16',  22-17'  and  17'-14'  require  to  be  counterbraced. 

Prob.  79.  A  deck  lattice  truss,  containing  four  web  sys- 
tems, like  Fig.  39,  has  20  panels,  each  12  feet  long,  the 
depth  of  t'-uss  is  24  feet;  the  dead  a.ul  live  loads  are 
500  lbs.,  and  1000  lbs.  per  foot  per  truss :  find  the  stresses 
in  all  the  members. 


Ill 


98 


HOOFS   AM)   ItniDGKS. 


V 


Alls.  Max.  stresses  in  upper  clionl  =z  18.0,  58.5,  94.5, 12G.0, 
iny.O,  ITo.u,  VJS.n,  L'Or.O,  lilO.O,  L'l'O.n  tons. 

Ma.\.  stresses  in  lower  cliord  =  lil'."),  (53.0,  UD.O,  i'M5 
157.5,  180.0,  198.0,  liU.S,  220.5,  225.0  tons.  ' 

Max.  stresses  in  struts  A-li,  2-5,  ete.  =  35.0,  31.8,  28.G, 
25.5,  22.7,  19.9,  17.2,  14.4,  12.1,  9.8,  7.4,  5.1,  3.2  tons. 

Mill.  stre.sses  in  struts  .1-3,  2-5,  etc.  =-11.7,  -10.(5, 
-9.5,  -S.5,  -7.0,  -5.5,  -4.1,  -2.0,  -0.8,  -f  1.2,  +3.2^ 
4-  5.1,  4-  7.4  tons. 

Max.  in  .1-1  =  22.5;  niin.  in  .1-1  =  7.5.  Max.  in  ^1-2 
=  <>7.5  ;  mill,  in  .1-2  =  22.5  tons. 

I'rob.  80.  xV  through  lattice  truss,  containing  four  web 
systems,  like  Fig.  39,  has  1»}  i)aiiels,  each  10  feet  long,  the 
(l.'].th  of  truss  is  20  feet;  the  dead  and  live  loads  are 
1000  lbs.,  and  2000  lbs.  per  foot  per  truss:  liud  the  stresses 
in  all  the  members. 

Art.  31.  The  Post  Truss.*  —  This  truss,  shown  in 
iMg.  40,  is  a  special  form  of  the  double  triangular  truss 
(Art.  28).     The  members  3-4,  5-0,  etc.,  are  struts,  and  all 


the  other  diagonals  are  ties;  the  counters  are  shown  in 
dotted  lines.  The  distinctive  feature  of  this  truss  is  that 
the  web  struts,  or  posts,  instead  of  standing  vertically,  are 

•  Tlxis  truss  has  becoiuj  ubsulote. 


BRIDGE  TliUSSES. 


99 


)4.6, 12G.0, 
!).0,   1^0.5, 

31.8,  •M.a, 

)ns. 

.7,  -  10.(5, 
1.2,  +  li.2, 

X.  in  ^1-2 


four  web 
:  long,  tlie 
loads  are 
le  stresses 


shown  in 
liar  truss 
s,  and  all 


shown  in 
ss  is  that 
cally,  are 


inclined  by  one  half  of  a  panel  length,  and  the  ties  by  one 
and  one  hulf  panel  lengths.  All  the  panels  of  Ixilh  chords 
aif  of  ctpud  length,  except  the  two  end  panels  oi"  the  lower 
chord,  which  are  each  one  half  a  panel  in  length. 

There  arc  and)iguities  in  the  character  of  the  stresses  in 
iliis  truss,  as  there  are  in  all  double  systems  of  bracing. 
It  is  impossible  to  separate  the  two  systems  as  they  are 
(connected  at  the  center.  But  if  we  assume  that,  under  a 
full  load,  the  members  7-10'  and  7-10  are  not  acting,  if  it 
is  a  deck  truss,  or  that  the  members  7-10',  7'-10,  9-10,  and 
9-10'  are  not  acting,  if  it  is  a  tlirough  truss,  tlien  the  chord 
stresses  are  readily  found.  Also,  if  we  assunu'  that  the 
systems  2,  1,  C,  5,  10,  7',  8',  3',  4',  1',  2',  and  2,  1,  4,  3,  8,  7, 
10',  ')',  0',  r,  2'  are  independent,  the  web  stresses  may  be 
readily  found. 

Prob.  81.  A  deck  Post  truss,  Fig.  40,  has  8  panels  in  the 
ujiper  chord,  each  10  feet  long,  and  20  feet  deep;  the  dead 
and  live  loads  are  400  lbs.,  and  IGOO  lbs.  per  foot  per  truss: 
lind  the  stresses  in  all  the  members. 

Dead  panel  load  per  truss  =  2  tons. 

Live  panel  load  per  truss  =  8  tons. 

tan  e  =  0.25,  and  sec  d  =  1.0308  for  the  posts. 

tan  fl  =  0.75,  and  sec  6  =  1.25  for  the  ties. 

Max.  stress  in  1-3=  [2  x  ^  +  1^  X  J]  10  =  16.3  tons. 

Max.  stress  in  4-6=  [2  x  J  +  2  x  i^]  10  =  10.0  tons. 

Max.  stress  in  5-6 

=  [Hx2+(2+44-6)]x  1.03=16.5  tons. 
Max.  stress  in  3-8 

=  [2  +  (1  4-  3  +  5)]  X  1.25  =  13.8  tons. 
Min.  stress  in  3-8=[2  -  8  x  i]  x  1.25  =  1.3  tons. 


iilf 
■  iii 

f 

im 
m 


i 


m 
if 


<etAi3t^»»ii>AT  KK 


KT 


% 


Ml 


"^''  nOOF&  AND  niilDGES. 

Thus  the  following  stresses  are  determined: 


Chord 

StHK!*8KS. 

Meuiikhs. 

i-a 

,S  .1 

28.7 

&-I 

;}rt.2 

7.2 

T-l) 

;i8.7 

7.7 

4-1 

10.0 
2.0 

c-t 

S  10 

lU  10' 

Max.  .    . 

Mill.  .    . 

25.0 
6.0 

a.-i.o 

7.11 

40.0 

8.0 

Sthkh.sivS  IX  Ties. 


Mkmiikks. 

14 

Ifi     1      8-8 

&-10 

7-10' 

7.6 
0.0 

S-'J 

I'-T 

4-:> 

Max.  .     . 
Mill.  .     . 

20.0 
4.2 

1S.7 
;!.8 

13.8 

8.7 
0.0 

5.0 
0.0 

2.-) 
0.0 

1..1 
0,0 

Stuk.sses  IV  Posts. 


Memiikrn. 

\-'i 

u-J 

.Ml 

7-8 

li-lo 

2-4 

Max.  .    , 
Mill.    .     . 

40.0 

8.0 

22.0 
4.2 

15.4 
.•i.l 

11.4 
2.1 

7.2 
1.0 

0.0 
0.0 

Prob.  82.  A  through  Post  truss,  like  Fig.  40,  lias  12 
panels  in  the  lower  cliord,  each  10  feet  long  and  20  feet 
deep ;  the  dead  and  live  loads  are  1000  ll,.s.,  and  2000  ll)s. 
per  foot  per  truss :  find  the  stresses  in  all  the  members. 

Art.  32.  The  sollman  Truss.  — This  truss,  shown  in 
I'Mg.  41,  was  the  earliest  type  of  iron  truss  built  in  the 
United  States.  It  eonsists  of  a  series  of  inverted  king  post 
trusses  with  une«pially  inclined  ties,  which  carry  the  load 
at  each  joint  directly  to  the  ends  of  the  upper  chord.  The 
vertical  pieces  are  struts;  the  upper  chord  is  in  compres- 
sion ;  and  there  is  uo  stress  in  the  lower  chord.     The  short 


■"■  *nJ>V.W**■"•^i^9«ll*fe^ii■.' 


fc,  'Jt^<^:^ij:^u-  <Hrft-i;^i»a*Wfe 


l''«^*4MS»«wasi4&'»b*w**rt6fl|^ftfci«!ti|^ 


i  10 

10  10' 

Ji.-i.O 
7.11 

40.0 

8.0 

(•-7 

4-:. 

2.". 
0.0 

1.3 
0.0 

1) 

2-4 

i 
3 

(».0 
0.0 

Mil 


40,  lias  12 
find  20  feet 
id  2000  lbs, 
embers. 

IS,  sliowu  in 
milt  in  tho 
d  king  post 
ry  the  load 
hold.  Tho 
ill  coinpies- 
The  short 


nniDCE  TRUSSES. 


101 


ties  shown  in  dolled  lines  in  each  panel  servo  to  stiffen  the 
structure. 


,1 

. 

. 

.V             1 

'^ 

s 

A 

0         a 

Vis.  41 

Holhnan  trusses  were  largely  used  in  the  Baltimore  and 
Ohio  Railroad  from  1840  to  18;"»0;  but  they  were  abandoned 
long  ago. 

Proh.  83.  A  deck  Uollman  truss,  Fig.  41,  has  0  jianels, 
each  12  feet  long,  and  IG  feet  deep;  the  dead  and  live  loads 
are  1(K)0  lbs.,  and  2000  lbs.  per  foot  per  truss:  find  the 
stresses  in  all  the  members. 

Art  33.  Tho  Fink  Trass.  — This  truss,  shown  in 
Fig.  42,  was  first  built  about  18r>l,  and  was  regarded  as  an 
improvement  on  the  Bollman  truss.     It  consists  of  a  com- 


A 

J      f 

;     n      K 

D' 

( 

;'     B'     A' 

^ 

5^ 

>. 

t 

\ 

^ 

^ 

:::^m 

a       h       e       (1       eb'     a' 

Fig.  43 

bination  of  inverted  king  post  trusses  with  equally  inclined 
ties,  as  the  primary  .system  AdA',  the  secondary  systems 
AhK  and  A'b'E,  and  the  tertiary  systems  AuC,  CcE,  etc. 
The  vertical  pieces  are  struts,  the  upper  chord  is  in  com- 
pression, and  there  is  no  stress  in  the  lower  chord ;  all  the 
diagonals  are  ties. 

Tiie  load  may  be  carried  either  upon  the  upper  or  the 
lower  chord,  though  it  has  been  used  more  often  for  deck 


V)2 


HOOFS  AM)  nnrnGEs. 


th.in  for  tluougli  l>ii(l(,'('3.  Tliesc  inisscs  were  built  from 
ISJI  down  to  about  l.S7<5;  but  they  arc  nut  now  generally 
regarded  with  favor  by  bridge  builders. 

I'riifi.  84.  A  deek  Kink  truss,  Fig.  41',  has  8  panels,  earh 
Ili  feet  long,  ami  Ki  feet  deep;  the  d.-atl  and  live  loads  are 
")(»()  lbs.,  and  2(>0()  lbs.  per  foot  per  truss:  find  the  stre.sses 
in  all  the  mend)ers. 

The  dead  panel  load  per  truss  = .'?  tojis. 

The  live  panel  load  per  truss  =  12  tons. 

Wo  see  at  onee,  from  Fig.  42,  that  the  maximum  stresses 
in  all  the  members  oeeur  when  there  is  a  full  panel  load  at 
every  apex.  Hence  the  maximum  stresses  will  be  when  1/5 
tons  arc  jdaeed  at  every  apex,  and  the  minimum  stresses 
will  be  one  fifth  of  the  maximum.  The  maxinnim  stress  in 
ea<'h  of  the  posts  lia,  Dc  is  —  15  tons.  Tin;  maximum 
stress  in  the  post  (Jb  is  the  panel  load  of  lo  tons  at  C,  plus 
one  half  the  panel  load  of  15  tons  at  />,  plus  one  half  the 
panel  load  of  15  tons  at  B,  =  —  30  tons.  Similarly  the 
stress  in  Ed  =  —  60  tons. 

Also, 

Stress  in  An  =--  ]  x  15  sec;  0  =  {).4  tons  =  On  =  Cc  =  Ec. 

Stress  in  Ah  =  |  x  ."iO  see  ",  =  27.0  tons  =  Eh. 

Stress  in  Ad  =  ^  X  00  see  0^  =  5)5.0  tons. 

Stress  in  AA'  =  -  [,J-  x  15  x  5  +  .J  X  ."0  X  2  +  J  X  CO  x  J^] 
=  -118.1  tons. 

Prob.  85.  A  deck  Fink  tntss  has  8  panels,  eaeh  10  feet 
long  and  15  feet  deep;  the  deid  and  live  loads  are  400  lbs. 
and  2000  lbs.  per  foot  per  truss:  tiad  the  stresses  in  all  the 
members. 


""^k*::ii,i^^A!%HMi&-J*'^i«i.^i 


i-i^»ir.^w.^\s*fc»#-,;j«rt^l!*«sa**t5i«tMjrt«ik.jM«Ui5afla'^ 


built  from 
ff  goneriilly 

)aneli),  each 
c  loads  are 
tho  striisst's 


nil  strosses 
lu'l  load  at 
l>o  when  1.5 
mi  stresses 
nil  stress  in 
maxiiiiuiii 
I  at  C,  plus 
le  half  the 
luilarly  th« 


=  Ec. 


X  60  X  J^] 

aeh  10  feet 
lie  400  lbs. 
s  ill  all  the 


HUWQK  TIlirssES. 


Rrid*!!'  Tkhshks  with  TNci.t\Ki»  OiinKns. 


108 


Art.  34.  The  Parabolic  Bowstring  Truss.  -lu  this 
truss,  Fig.  l.'{,  tiie  lower  ulioiil  is  horizontal,  ami  the  uiijier 


^-'r<rixlM7Pt>r^ 


A  EI)      C 

yilu  Via.  *3 


D 


chord  joints  lie  on  the  arc  of  a  parabola ;  the  verticals  may 
be  in  compression  and  the  diagonals  in  tension,  or  tlie  ver- 
ticals may  be  in  tension  and  the  diagonals  in  compression. 

Let  I  =  the  length  of  the  span  vl/i,  d  =  the  height  of  the 
are  (>C  at  the  center  of  the  sjian,  and  w  =  the  uiiiforiii  load 
in  lbs.  ]ier  foot  of  truss;  let  (.r, y)  be  any  joint  /'of  the 
ujiper  chord  referred  t'-  the  axes  ATi  and  AY,  and  S  the 
stress  in  the  lower  chord  panel  opposite  P.     Then  we  have 


(1) 


The  equation  of  the  parabola  referred  to  its  vertex  O  as 
origin  is 

{\l-xf^2p{a-v) (2) 


72 

and  for  the  point  A  this  becomes  —  =  2pd\ 
Sub.stituting  this  in  (2)  and  solving  for  y,  we  get 

y=^^Hx-aP)     .     .     .     . 


2/)  = 


All 


which  in  (1)  gives 


5  = 


(3) 
(4) 


IIpncp,f(>r  a  uniform  loud  on  ii  pnrdhoUc  truss  the  stress  in 
the  lower  chord  is  the  same  in  every  panel. 


■ ' '  u:iavtm^f^i.  S"'^*"' 


104 


IKXIFS   A.\h   nitlDCKs. 


Sim-e  llu-  liorizoiital  compoiiPiit  of  flip  stress  in  any 
iliaj,'<)ii:il,  iis  /'/>  for  cxainiilr,  is  f(|iial  to  the  (lillfiviiti-  ot 
tlio  diord  iitri'sscH  in  the  adjiiceiit  iiaiiols,  /<;/>  and  IX'. 

Thvi-cf()ri>,  fitf  (I  Kiiijhnn  loiul  there  ix  uo  .slretrn  in  an  if 
dkiyuiKil. 

It  fuUowH  at  once  tliat, /or  ti  iDiifhrm  load  tho  hitrizautnl 
comjinni'Ht  of  the  .s/ms.s  i„  the  n/iiter  churd  in  the  name  in  erery 
imnel,  and  is  expri'sstul  by  ('(iiiation  (4j. 

Therefore,  for  a  intifnnn  Indd  the  stress  in  nvij  inrlined 
jntnel  of  the  np/ier  chord  is  eiimd  to  the  stress  in  a  fninel  of  the 
lower  chord  mnltijtiieil  lij/  the  secant  of  the  inclination. 

I'roli.  86.  A  paralKjlic!  liowstrini,',  Kij,'.  11,  as  a  tlir(ni},'li 
biidgt;,  has  8  panels,  oaeli  1(»  feet  loiiy,  and  lU  feet  eenler 


Kin.  44. 

depth,  the  verticals  t;>ke  either  tensioji  or  eompression,  the 
diai,'onals  are  ties;  the  dead  and  live  loads  are  4(10  lbs.,  and 
S(M>  lbs.   per   foot  per  truss:    hnd    the  stre.s.ses  in    all    the 
niendters. 
The  max.  .stre.ss  in  any  panel  of  the  lower  chord,  by  (4), 

1200  X  .so-'      ,„  ^ 

.-.  Min.  stress  =1  x  48  =  IG  tons. 

Dead  panel  load  =  li  tons. 
Live  ])anel  load  =  4  tons. 


^^.^^.^^<^Mw^A.ca»^saw^^,^i^;y^:.^^a=lrit^ 


■Jk 


T 


II  It  I  in;  K   TltUHS.ZH. 


lor) 


%  (.">)  we  iiiivc  lfiit,'tli.s  of  ;{-J,  ;"■)(;,  iind  7  S  =  L.'JTr.,  7.r>, 
ami  l)..'{7r»  ft't't. 


Max.  stress  in  m_;,  =  48_x y  (1  ()) ^  f  (1;{7;^^ 


II) 


=  fi2.4  tons. 


Sinco  the  (load  load  pnidiicis  no  .stroHscs  in  tlic  diaKonals, 
the  max.  stress  in  any  diagonal  is  found  liy  |iiittinH  only  tin- 
live  load  on  tho  hrid^,'o  in  the  position  to  },'ive  the  max. 
shear  in  that  memlier  (Art.  '2'J,). 

Thus,  for  the  maxinuim  stress  in  H-S  the  live  load  is 
l»liu;ed  on  the  right,  the  center  of  moments  is  on  tho  lower 
chord  at  L'(»  feet  to  the  left  of  the  i)oint  2.  The  lever  arm 
of  r.  8  is  ;!(►  feet. 

The  reaction  for  this  loading 

=  1(1  +  2  +  3  +  4  +  5)  =  7.fi  tons. 
.-.  Stress  in  5-8  =  ^^^  =  5  tons. 

till 

For  live  load  on  the  left,  the  stress  in  5-8  =  0,  and  the 
connter  0-7  comes  into  action.     The  lever  arm  of  (5-7  is 
27.4  feet ;  tlie  reaction  for  this  loading  =  (i.a  tons.     There- 
fore the  stress  -S'  for  G-7  is  found  from  the  eciuatiou 
G.5  X  20  -  4  (30  4-  40)+  .S'  x  27.4  =  0. 
.-.  S  =  stress  in  0-7  =  5  /»  tons. 
If  we  cut  a  vertical  as  5-(i  by  a  section,  place  the  live 
load  on  tho  right,  and  take  the  center  of  moments  at  tho 
intersection  of  .'}-,'>  and  4-0,  which  is  4  feet  to  the  left  of 
the  point  2,  we  shall  obtain  tlie  maximvim  eompresslon  iu  it 
caused  liy  the  live  load.     Thus, 

The  reaction  for  this  loading  =  71)  tons. 
■  •.  Max.  stress  in  r)-(>  due  to  live  load 
7.5  X  4 


24 


■  =  —  1 .25  tons. 


Min.  stress  in  5-0  =  2.0  -  1.25  =  0.75 


tons. 


I! 

11 


i 


100 


/•OOFS  Axif  iiiiihaKS. 


Th«  tiiii.xiiimm  HtrcsH  in  cm-h  vorticul  is  omo  of  tciisidii, 
aiitl  will  (MTiir  wlifii  \\w  livo  Iduil  fovors  tlio  hritlgu,  ami  is  u 
(loiul  ami  livo  piiiicl  l((iul,  nr  ()  tons. 

It  llicrc  bt!  IK)  live  Idiul  dm  tlit'  tiiiss,  fho  strtss  in  any 
vcrtiful  is  u  dead  imni'l  load,  or  a  tt'nHhm  of  '*  toii.s.  Now 
let  till'  Iivt(  load  conn,  on  from  tli«  rij,'lit,  a  iiart  of  it  which 
KOCH  to  the  left  support  will  cuii.sc  ruiiiprr.s.sion  in  each 
vertical,  or  will  diminish  the  tension  in  that  vertical  duts 
to  the  .load  load.  Thus,  when  the  live  load  reaches  the 
joint  8',  it  will  cause  in  the  vertical  U-IO  a  compression  of 
'J.'J  tons,  which  will  neutralize  the  dead  loatl  tension  of 
2  tons,  and  leave  as  the  result  a  compression  of  0.1'  tons. 
Also,  when  it  reaches  the  ape.K  S,  it  will  cause  in  the  vortical 
fi-(;  a  compression  of  l.L'o  tons,  as  we  .saw  above,  which  will 
neutralize  that  much  of  the  dead  loail  tension  of  li  tons,  .-.nd 
leave  as  the  result  a  tetision  of  0.7ri  tons. 

'i'h(i  following  stresses  are  found  in  a  manner  similar  to 
the  above ; 

Ciioni)  Stiikhsks. 


Mkmiikkh. 

Max.  .  .  . 

-62.4 
-  17.5 

.'t  ft 

f-7 

•-9 

-48.1 

-  irt.o 

•i-i,  4-1!,  otc. 

+  48,0 
+ 10.0 

-60.:. 
-16.8 

-48.8 
-  WM 

Strk«se8  in  Diaoonals, 


MKMIl'Jh.S. 

3-0 

.vs 

7-10 

■1-,^ 

8  7 

S-9 

Max.   .  .  . 
Min.    .  .  . 

+4.;} 

0.0 

+  5.0 
0.0 

+  5.6 
0.0 

+6.0 
0.0 

+  5.5 
0.0 

-^5.7 
0.0 

■■>»-a«.j»,s**.5ii»9*Mi-te'»»M,.t>.wSjSi«i;jj(isj(Miwiiiwaiai^ 


>f    t<MlHi(lll, 

;u,  uiul  i.s  ii 

cM.s  in  any 
)n.s.  Now 
f  it  wliich 
I  ill  t'iicli 
rlical  (lilt! 
'iiclics  tli«' 
•iCHsion  of 
fi'ii.sion  of 
F  0.1'  toiiH. 
\\G  vpiticiU 
vliicli  will 
ton.s,  ■.'.\n\ 

siinilar  to 


i-l,  4-fl,  olc. 

+48.0 
+ 1(1.0 


s-9 


-^5.7 
0.0 


nninfiK  thussksi. 


Sthkhxk*  in  Vkiitic»i.«. 


KH 


Mr.MHrHA. 

+0.0 

0.0 

,'Ml 

■-S 

»  III 

.Max 

+6.0 

+  0.h 

+6.0 

+  0.0 

+  0.0 
-  0.2 

Mill 

I'rnh,  87.  A  pamlMiliir  howstrinj,'  a.s  a  tlirnii^li  brid^O) 
Fig.  44,  has  H  punols.  t'iirli  10  foct  long  and  10  feet  center 
(i('i)tli;  till!  verticals  ar«  tics  and  (lie  diaifonalH  arc  braces; 
tlic  (lead  and  live  loads  are  100  lbs.  and  1(100  jhs.  per  foot 
|»c''  truss;  find  the  stresses  in  all  the  meiiibers. 

Alls.    Max.  stress  in  each  panel  oi'  lower  (•liord=  +80  tons. 
Ma.x.  stresses  in  iijiper  chord 

=  -  S7.;5,  -  S.'{.S,  -  81.4,  -  80.L'  tons. 
Max.  stress  in  .■!-(» 

=  -  8.7,  in  r)-8  =  -  10.0,  in  7-10  =  -  10.'),  in  4-5 
=  -  10.0,  in  (h-7  =  -  10.'.),  in  8  '»  =  -  ll..'i  tons. 
Max.  stresses  in  verti(^als 

=  +-10.0,  +VJ.r>,  +14.0,  +-14.5  tons. 
Mill,  stresses  in  verticals^ +2.0,  +2.0,  +-2.0,  +2.0  tons. 
Mill,  stres.ses  in  diagonals  =  0. 

J'liil).  88.  A  thidiigli  paraboli(!  bowstring  truss,  Fig.  44, 
has  8  panels,  eaidi  12  b-et  long  and  IG  feet  center  depth  ;  the 
verticals  arctics  and  the  diagonals  are  braces,  the  dead  and 
live  loads  are  500  lbs.  and  2000  lbs.  per  foot  per  truss:  liiid 
the  stresses  in  all  the  menibers. 

Alls.    Max.  stress  in  eacdi  panel  of  lower  chord  =  +  JM)  tons. 

Min.  stress  in  each  pamd  of  lower  chord  =  +- 18  tons. 

Max.  stres.ses  in  upper  chord 

=5  -  104.1,  -  D7.5,  -  1)2.7,  -  90.3  tons. 


wfffWBlK^T'y**^ 


mmm 


iJk 


108 


HOOFS  AND  liltlDGES. 


r.Tiii.  stresses  in  upper  cliord 

=  -  L'()..S,  -  V.).r>,  -  IS.n,  - 18.1  tons. 
Max.  stresses  in  verticals 

=  +  ir).0,  +18.7,   f-21.0,  +21.7  tons. 
Min.  stresses  i"  verticals  = +3.0,  +3.0,  +3.0,  +3.0  tons. 
Max.  stress  in  3-(> 

=  -  10.4,  m  5-^  =  -  12.7,  in  7-10  =  -  M.4,  in  4-5 

=  -  12.7,  in  G-7  =  -  14.4,  in  8-9  =  -  15.0  tons. 
Min.  stresses  in  diagonals  =  0. 

Pmb.  89.  A  tliroii.L,']]  iiarabolic  bowstring  truss  has  12 
panels,  eac'i  S  Jeet  long,  ami  12  feet  center  depth;  Ihe 
verticals  are  ties  and  the  diagonals  are  braces;  the  dead 
and  live  loads  aie  oOO  lbs.  and  2000  lbs.  per  foot  per  truss: 
tind  the  stresses  in  all  the  members. 

Art.  35  The  Circular  Bowstring  Truss.  — This  form 
of  tiuss,  Fig.  45,  is  often  used  for  highway  bridges.     'l"he 


joints  of  the  upper  chord  lie  upon  the  arc  of  a  circle,  each 
joint  being  directly  over  the  middle  of  the  panel  below. 
The  diagonals  are  built  to  take  either  tension  or  com- 
pression. 

Proh.  90.  A  through  circular  bowstring  truss,  Fig.  45,  has 
G  panels,  each  12  feet  long,  and  11.7  feet  center  depth ;  the 
joints  of  the  upper  chord  lie  on  the  arc  of  a  I'ircle  of  00  feet 
radius;  the  dead  and  live  ionils  mv  150  li)s.  and  I.'tOO  lbs. 
per  foot  per  truss:  find  the  stresses  in  all  '1  e  uiend)ers. 


•*i««aiW««wiw*«i7W«a»«t!i&-i«e^^ 


1 


HI! ma!':  thusnes. 


10!) 


A)tH.  Max.  stress  in  '_'-.'{=.- 40.4,  in  3-5  =  - 54.3,  in 
5-7  =  -  51.1,   in  7-7'  =  -  50.1  tons. 

Ma.\.  stress  in  2-4  =  +  41.1,  in  4-6  =  +  45.8,  in  fi-8 
=  +  47.3  tons. 

Max.  stress  in  3-4  =  +  9.0,  in  4-5  =  +  9.0,  in  5-(i 
=  +  10.4,    in  G-7  =  +  10.0,   in  7-S  .■=  +  10.8  tons. 

Min.  stress  in  4-5  =— 2.2,  in  5-6=— 1.3,  in  0-7 
=  -3.0,   in  7-8  =  -3.1  tons. 

This  pnibi'Mii  is  lakcn,  with  slight  changes,  from  Burr's  Stresses 
in  nnilj;e  and  IJoof  Triussos. 

Proh.  91.  A  throu,t,'h  circular  Oow.striiif;  truss  has  8 
panels,  eaoh  15  feet  long,  the  center  depth  is  10.74  feet; 
the  joints  of  the  upper  chord  lie  on  the  are  of  a  circle  of 
100  feet  radius,  each  joint  being  directly  over  the  center 
of  the  panel  below;  the  dead  and  live  loads  are  KKKt  lbs., 
and  2000  lbs.  per  foot  per  truss:  find  the  stresses  in  all  the 
members. 

Art.  36.  Snow  Load  Stresses.  —  For  railway  bridges 
the  snow  load  is  not  taUen  into  account,  since  the  floor  is 
opeUi  so  that  but  little  is  retained.  For  highway  bridges 
the  snow  load  is  taken  from  0  to  15  lbs.  per  square  foot  of 
floor  surface,  depending  upon  the  climate  where  the  bridge 
is  situated.*  The  snow  load  is  tcaken  lower  than  for  roofs, 
since  the  full  live  load  is  not  likely  to  come  upon  the  bridge 
v.'hile  it  is  heavily  loaded  with  snow.  Since  the  snow  load 
is  uniform,  the  stresses  due  to  it  are  computed  in  the  same 
way  as  the  dead  load  stresses ;  or  the  snow  load  and  dead 
load  stresses  are  proportional  to  the  corresponding  apex 
loads  (Art.  8). 

*  In  bnilding  highway  bridgen  iu  England  and  F  rauce  the  snow  load  is 
not  generally  considered. 


ii 

!; 

i  ; 
!!) 
if 


,./■ 


no 


HOOFS  A\J}   lUUDdES. 


i 


Pruh.  92.  A  thioufijli  llowo  truss,  Fig.  32,  him  10  icuiels, 
each  12  feet  long  ami  12  tVct  deoj);  tlie  width  of  roadway 
is  20  feet,  and  the  width  .  eiudi  sidewallc  i.s  ;"»  feet;  tlic 
snow  h>ad  is  10  lbs.  per  stjiiare  foot  of  floor :  find  the  snow 
load  stresses  in  all  the  niendiers. 

Ann.   Stresses  in  lower  chord 


=  4.1, 


9.5,   lO.S,  1  l.;j  tons. 


Stresses  in  the  verticals  =  4.1,  3.2,  2.3,  1.4,  0.9  tons. 
Stresses  in  the  diagonals 

=  -S.7,  -4.4,  -3.2,  -.9,  -lO.Gtons. 

P)-(>b.  93.  A  (leek  Pratt  truss.  Fig.  31,  has  10  jtanels,  eaeh 
10  feet  long  and  1(5  feet  deep;  the  width  of  roadway,  in- 
cluding sidewalks,  is  3o  feet;  the  snow  load  is  15  lbs.  per 
square  foot  of  floor:  find  the  snow  load  stresses  in  all  tlie 
members. 

Art.  37.  Stresses  due  to  Wind  Pressure.  —  In  bridges 
of  large  sjjan,  it  is  ofleu  found  that  the  stresses  produced  in 
some  of  the  members  by  a  gale  of  wind  are  almost  as  great 
as  those  caused  by  both  the  dead  and  live  loads.  In  the 
principal  members  of  the  Forth  bridge,  Scotland,  Sir  15en- 
jamin  Maker  estimated  that  tlu?  maximum  stresses  due  to 
these  t'iiree  separate  forces  were  as  follows : 

Stress  due  to  dead  load  =  2282  tons. 
Stress  due  to  live  load  =  1022  tons. 
Stress  due  to  wind  Joad  =  2920  tons. 

In  estimating  the  wiiul  pressure,  ami  the  resulting  stresses 
in  the  mend)ers  of  a  bridge,  the  practice  of  engineers  varies 
greatly.  Different  railroads  have  their  own  speciiications 
for  wind  i)rcssurc.  The  standard  wind  ])ressure  i)er  scpiarc 
foot  ranges  in  this  country  from  30  to  50  lbs.  It  is  assumed 
by  many  engineers  that  there  will  be  no  train  on  the  bridge 


^?*3<iil*l««>»l^i,^»^,•«»4iWMM«Ba»**K■<*.  - 


lilllDGE  TltUSSkS. 


Ill 


whiiii  the  wind  hli.ws  with  greater  pressure  than  .30  lbs.  per 
S(|nare  foot.  A  wind  pressure  of  30  lbs.  per  square  foot  will 
overturn  an  empty  freigiit  car.  In  sucli  a  gale  it  would  be 
hardly  possible  for  a  train  to  reach  the  bridge.  It  might, 
however,  be  eaught  there  by  a  sudden  s(piall ;  and  this  is 
just  what  ajipi-ars  to  hav(!  happeiu'd  in  the  case  of  the  Tay 
bri<lge,  whicdi  was  destroyed  by  a  wind  storm  in  1S7<.>.  A 
maximum  pressure  of  oO  lbs.  per  square  foot  is  taken  as 
applying  to  the  bridge  alone. 

The  surface  exposed  to  the  wind  action  is  commonly 
taken  as  ilouble  the  side  elevation  of  one  truss,  for  the 
vca.son  that  the  windward  truss  cannot  afford  much  siielter 
to  the  leeward  truss,  whatever  may  be  the  direction  of  the 
wind.  In  a  heavy  gale  of  wind  there  is  not  much  shelter  to 
be  fouiul  ujider  the  lee  of  a  lamjvpost  at  a  distance  of 
lit)  feet  from  it,  even  if  the  post  be  directly  to  windward. 
Experiments  show  that  the  wind  pressure  against  the  two 
trusses  of  a  bridge  is  more  than  1.8  times  that  on  the 
exjwsed  surface  of  one  truss. 

For  Highway  Bridges  the  wind  i)ressure  is  fretpiently 
taken  at  about  30  lbs.  per  square  foot  of  exposed  surface  of 
both  trusses.  It  is  assumed  that  there  will  be  no  live  load 
upon  the  bridge  when  the  wind  is  blowing  at  this  maximum 
pressure  of  30  lbs.  i)er  squar"  foot. 

For  Railroad  Bridges  tlie  wind  pressure  is  taken  at 
about  30  lbs.  per  scpiare  foot  of  exposed  su  face  of  bv)th 
trusses,  and  about  3(K>  lbs.  per  linear  foot  due  to  the  train 
surface.  The  train  surface  is  about  10  square  feet  for  each 
linear  foot  of  the  bridge.  The  30  lbs.  per  square  foot  of  the 
trusF^s  is  treated  usually  as  a  dead  load,  and  the  300  lbs. 
per  linear  foot  flue  to  the  train  surface  as  a  live  load.  This 
regards  ejich  truss  as  fully  exposed  even  when  the  train  is 
on,  though  it  partially  shelters  one  t;  uss. 


112 


UOOFS  Ay  I)   niilDGES. 


Ti)  estimate  the  .'50  lbs.  pressure  \iev  s<iuare  foot  of  exposed 
surface  of  both  trusses,  wlien  this  surface  is  not  known, 
it  is  eustonuiry  now  to  us»  the  followini;  nth':  Tiikr  loO 
//as.  per  linear  foot  per  triisfi,  or  75  lbs.  per  linen  r  fool  for  emih 
chord. 

The  .vind  ioad  on  the  trusses  is  assumed  to  be  diviiU'd 
equally  between  the  upper  and  lower  lateral  trusses,  while 
the  wind  load  on  the  train  is  all  taken  by  the  lateral  truss 
belonging  to  the  loaded  chord.  The  lateral  trusses  are  the 
iKU'izontal  trusses  placed  between  the  chords  of  the  vertieal 
trusses.  The  chords  of  tin;  vertical  trusses  are  also  the 
chords  of  the  lateral  trusses.  The  lateral  trusses  of  a  bridge 
are  either  of  the  Pratt,  IFowe,  or  Warren  type,  correspond- 
ing geiH'rally  with  the  type  of  the  main  trusses. 

IJkm.  —  111  tiiuliii^  wiiul  atrt'sses  th»i  loads  on  tlii'  laiTal  system 
of  the  uuloadc  1  chord  are  considered  as  applied  equally  iiixm  the  two 
sidoH,  winihvard  and  leewanl,  while  the  loads  on  the  hiteral  sysletn  of 
the  loaded  chord  are  considered  iw  applied  wholly  on  the  windward 
side.  The  stres.ses  are  then  readily  found  by  methods  already  familiar 
for  liiidin^'  dead  load  anil  live  load  stresses. 

The  lower  lateral  system  in  a  througL  bridge  and  the  upper  lateral 
system  in  a  ileck  bridge  are  calculated  for  a  dead  load  of  ;50  lbs.  per 
.si[iiare  fool  of  exposed  surface  of  both  trusses,  and  a  live  load  of 
."JOO  lbs.  per  linear  foot  of  train  ;  while  the  other  lateral  .system  is 
calculated  only  for  a  dead  load  of  30  lbs.  per  square  fool  of  exposed 
surface  of  both  trus.sos. 

The  resulting  chord  stresses  should  be  combined  with 
tlu)se  diu!  to  the  dead  load,  or  with  those  due  to  the  dead 
and  live  loads  if  the  live  load  acts  with  the  wind  load.  The 
lower  chord  is  always  in  tension  under  the  action  of  the 
dead  and  live  loads.  When  the  wind  acts  the  bridge  is  bent 
laterally,  and  the  windward  lower  chord  has  its  maximum 
tension  diminished  by  the  compression  due  to  the  wind, 


■  -hiilftiilSaaieMitt^^tiaMiiteJfCjait,,. 


of  exposed 

lot  known, 

T,ih-  ir.(» 

tiol  for  iiu'h 

lie  divided 
isses,  while 
iteral  trusn 
3es  are  the 
the  vertical 
re  also  the 
of  a  bridge 
correspond- 


ii?ral  Rystem 
H|)(in  the  two 
r.il  sysli'iii  of 
lu!  wiiidwanl 
L'iu'.y  familiar 

upper  lateral 
i)f  ."JO  lbs.  per 
live  load  of 
ral  system  is 
)t  of  exposed 


bined  with 
()  the  dead 
load.  The 
:ion  of  the 
idge  is  bent 
niaxiniiiMi 
the  wind, 


BRIDGE  TRUSSES. 


113 


•while  the  lower  leeward  chord  has  its  tension  iiKTeased  by 
the  tension  due  to  the  wind.  The  compression  in  the  wind- 
ward chord  due  to  the  wind  may  exceed  the  tension  due  to 
the  dead  load,  or  to  the  dead  and  live  loads  condiined.  The 
lower  chord  on  each  side  then  shoii'd  not  only  be  able  to 
sustain  the  total  maximum  of  dead,  live,  and  wind  loads,  but 
to  act  as  a  strut  to  jesist  compression. 

I'roh.  94.  A  through  Vratt  triiss  railw.iy  bridge  16  feet 
wide,  half  of  which  is  shown  in  Fig.  47,  has  12  panels,  ^jach 
IG  feet  long;  the  upper  and  lower  lateral  systems,  Figs.  46 


VD 

C 

(i 

e 

/ 

A 

\^y 

\/ 

\  ^'' 

\, 

•    x 

/     \ 

/\ 

/    X. 

/                Xj 

/ 

b-     c'      a'       6'      /'       A' 

Ficr.  46 


b         c 

<t          « 

/ 

* 

/ 

\       / 

\        /' 
\     / 

\            / 
\      / 

\      / 

«'v 

/ 

/  \ 

/   \ 

/      \ 

/   \ 
/      \ 

/          \ 

/     \ 

/ 

Fig.  47 


and  4S,  are  Pratt  trusses :  tind  the  stresses  in  both  lateral 
systems  due  to  a  wind  pressure  of  .'{()  lbs.  i)er  square  foot  of 
ex|)osed  surface  of  bo'h  trusses  and  300  lbs.  per  linear  foot 
of  train  surface. 


in 


ROOFS   AND   nniDUBS. 


(1)    Ui'i'KR  Latkhai,  Systkm  (10  I'ankls). 


Panel  load  for  one  chord  (see  rule)=  "'.^  ."'  =  O.Ci  tons. 


To  X  1<> 
L'OOO 

This  load  acts  as  a  dead  load  at  each  apex  of  the  wind- 
ward and  leeward  chords  in  the  direction  of  the  arrows, 
shown  in  Fij^.  H'r,  hence  there  is  no  stress  in  the  dotted 
diai^onals. 

tantf=l;    sec  5  =  1.414. 

Stress  in  hh'  =  —  0.6  tons. 
Stress  in  //''  =:  -  O.G  x  2  =  -  1.2  tons. 
Stres.s  in  ce'  =  —  0.(>  x  4  =  —  2.4  tons. 
Stress  in  he  —  —  ().<»  x  1)  x  1  =  —  0.4  ton 
Stress  in  al  =  -  [o.4  -f-  4.2]  x  1  =  -  9.0  tons. 
Stress  in  fh'  =  -f-  0.0  x  1.414  =  4-  O.S  tons. 
Stress  in  c/'  =  +  3  x  O.G  x  1.414  =  -f  2.5  tons. 


(2)   Lower  Lateral  System  (12  Panels). 

Panel  load  for  both  chords 
ino  X  10  ,  300  X  10 


■  + 


=  1.2  tons  +  2.4  tons. 


2000  2000 

The  former  we  treat  as  a  dead  load,  the  latter  as  a  live 
load,  both  acting  at  each  ajicx  of  the  windward  chord  in  the 
direction  of  the  arrows,  shown  in  Fig.  48.     (See  Remark.) 

Stress  in  AB  =  -  6.6  x  3.0  =  -  19.8  tons. 
Stress  in  C't" 


:-r4.6xl.2+|i(l-f2+3-h  -  +10)1= -16.4 


tons. 


Stress  in  DE' 

=  [2.ox  1.2+0.2(1+2+3+  -.  +8)]  1.414=  14.4  tons. 


L8). 

:  ().(■»  tons. 

f  the  wind- 
;lio  arrows, 
the  dotted 


HIillK.K  TIlUiiSKS. 

TliiLs  the  I'ullowiiij;  strcssos  are  (M)iiiimted: 

ITl'I'KIt    liATKKAl-    SVSTKM. 

Stresses  in  chords 

=  -  r>A,  - <».(;,  -  V2.(\,  -  It. 4,  -  m.o  tons. 

Stresses  in  struts 

=  -  r>.7,  -  l.H,  -  :U\,  -  'lA,  -  l.L',  -  0.0  tons. 
Stresses  in  dia^jonals 

=  -f  7.(i,  +").•.»,  +  1.1,',  -+!-'.'»,  +0.8  tons. 


11  ■> 


)ns. 


ons. 


IS). 


ns. 


I"  as  a  live 
liord  in  the 
Remark.) 


16.4  tons. 


L4.4  tons. 


LoWEK   LATKIlAr,   Sv.STEM. 

Stres.ses  in  chords  -^  10.8,  30.0,  48.G,  .57.6,  G.3.0,  G4.8  tons. 
Stresses  in  struts 

=  -'2\.(\,  -1J).8,  -10.4,  -i;i.2,  -10.2,  -7.4,  -ratons. 
Stresses  in  diagonals  =  27.9,  23.1,  IS.O,  11.4,  10.4, 0.8  tons. 

Tiie  cliord  bciUi  is  in  eompression  under  the  action  of  the 
dead  and  live  loads;  this  conii»ression  is  increased  by  that 
due  to  the  wind  pressure,  as  just  found. 

The  chord  A'WC'Il'  is  in  tension  under  the  action  of  the 
dead  and  live  loads ;  this  tension  is  increased  by  that  due  to 
the  wind  pressure,  as  just  found. 

When  the  wind  Idows  on  the  opposite  side  of  the  bridge 
the  diagonal  and  chord  stresses  are  to  be  interchanged, 
while  the  strut  tresses  remain  the  same.  This  there  will 
be  the  same  stresses  in  b'c\  c'd',  etc.,  as  in  Ix',  cd,  etc.,  and 
the  same  in  A'li',  Ti'C,  etc.,  as  in  Ali,  li<\  etc.,  and  the 
dotted  system  of  braces  will  act  in  eiuOi  lateral. 

Tlie  wind  huvds  on  the  left  hidf  of  the  upper  lateral  are 
transicrred  to  the  abutment  at  A  by  means  of  the  portal 
hrac,':'tj  AA'b'b  in  the  transver.se  plane  of  Ab. 


no 


/?OOF.S   AST)  niilDQES. 


Proh.  96,  A  tliruugli  IIowo  truss  niilwuy  briiltjo  -<>  f't'ct 
wide  has  I'J  paiu'ls,  cacli  '-'()  U'ci  loiif,';  llic  upper  anil  lover 
lateral  systems  aro  Howe  trusses:  iind  the  stresses  in  Ix.th 
lateral  systems  tlue  to  the  same  wind  pressure  as  in  tiie 
previous  problem. 

Hy  taking  the  dotted  diagonals,  Fij,'s.  47,  M!,  and  48  will 
represcMit  tlie  left  half  of  the  Howe  truss  and  the  uppe- and 
lower  lateral  systems.     Then  by  riil<' : 

i'aiiel  load  for  one  chord  of  upper  lateral  =  0.75  tons,  to 
he  taken  as  a  dead  load. 

I'anel  load  for  both  ehords  of  lower  lateral 

=  1.5  tons  +  .'{.0  tons ; 

tho  former  to  be  taken  as  a  dead  load,  the  latter  as  a  live 
loa?^;  all  acting  in  tho  direction  of  the  arro  vs. 
Stress  in  b'c  =  -  9  X  .75  X  1.414  =  -  9.5  tons. 

Stress  in  CD 
=  -  [3.5  X  1.5  4-  -rSj(l  +  2  +  ■■•  +  9)]  1.414  =  -  23.3  tons. 

Stress  in  B'C  =  10  x  4.5  x  1  =  45  tons. 

Upper  Lateral  System. 
Stresses  in  chor.ls  =  G.S,  12.0,  15.8,  18.0,  18.8  tons. 
Stresses  in  struts  =  0.4,  G.O,  4.5,  3.0,  1.5,  0.8  tons. 

Stresses  in  diagonals 

=  _  9.5,  _  7.4,  -  5.3,  -  3.2,  -  1.1  tons. 

LowKK  Lateral  System. 

Stresses  in  chords  =  24.8,  45.0,  60.8,  72.0.  78.8,  81.0  tons. 

Stresses  in  struts  =  2..3,  20.5,  1G.5,  12.8,  9.2,  6.0,  3.8  tons. 

Stresses  in  diagonals 

=  _  34,9,  _  28.9,  -  23.3,  -  18.0,  -  13.0,  -  8,5  tons. 


ii 


iiIk'o  L'O  fVct 
r  anil  lover 
;ses  ill  l)(>t)i 
0  as  ill  tlio 

and  48  will 
e  vippe-  ami 

).75  tons,  to 


cr  as  a  live 


-  23.3  tons. 


tons. 
;on8. 


I,  81.0  tons. 
5.0,  3.8  tons. 

■  8.5  tons. 


hltlDGE  riiVssKs. 


\V 


I'rttli.  96.  A  ilock  Pratt  truss  railroad  briilj,'*-  -<>  fi't't  wido 
has  10  panels,  each  i'O  feet  long ;  the  upixT  and  lower  lateral 
systems  are  Pratt  trusses:  'ind  the  stresses  in  both  lateral 
systems  dno  to  a  wind  pressnre  of  40  Ihs.  |.er  stpiare  foot  of 
expo.sed  surface  of  lioth  trusses  and  3(M)  Ihs.  i)er  linear  foot 
of  train  surface. 

Hy  taking  only  the  fidl  diagonals,  Fig.  4('»  will  represent 
the  left  half  of  the  deck  Pratt  truss  and  also  the  upper  and 
lower  lateral  systems. 

Since  this  is  a  deck  bridge,  tlie  upper  Iriterals  are  on  the 
loaded,  and  the  lower  laterals  are  on  the  unloaded  chord. 
Hence  by  the  rule: 

I'anel  load  for  one  eliord  on  lower  lateral  system 

20()0     ' 


1  ton. 


to  be  taken  as  a  dead  load  at  each  apex  of  the  windward 
and  leeward  chords  (Rem.). 

Panel  load  for  both  chords  on  upper  lateral  system 
200  X  20  ,  300  X  20      „  ,         ,  .,  , 
=  "2000-  +  ~-2mr  =  ^  ^''"^  + ''  ^""« ' 

the  former  to  be  taken  as  a  dead  load,  the  latter  as  a  live 
load,  both  acting  at  each  apex  of  the  >vitidward  chord  (Item.), 
and  all  acting  in  the  direction  of  the  airows  in  Fig.  40. 

Uri'KU  Lateral  System. 

Stresses  in  chords 

=  -  22.5,  -  40.0,  -  52.5,  -  GO.O,  -  62.5  tons. 
Stresses  in  struts 

=  -  25.0,  -  22.6,  -  17.8,  - 13.4,  -  9.3,  -  5.0  tons. 
Stresses  in  diagonals 

=  +  31.7,  +26.1,  +18.9,  +  13.1,  +7.7  tons. 


cr 


118 


HOOFS  AM)  niilDOES. 


LoWKH    liATUKAL   SvHTKM. 

Stresses  in  chonls 

=  -9.0,  -10.0,  -21.0,  -24.0,  -25.0  tons. 
Stri's.st's  in  struts 

=  -  9.5,  -  HA),  -  CO,  -  4.0,  -  2.0,  -  1.0  tons. 
Stresses  in  (Ha^jonals 

=  +  12.7,  +  1»-1».  +  7.1,  +  4.2,  4- 1.4  tons. 

Prof).  97.  A  throiigli  I'ratt  tni.ss  railroad  bridge  IfiJ  fett 
wido  lias  0  paticl.s,  cacli  17  f»'ot  long;  the  upper  and  lower 
lateral  systems  are  Tratt  trusses:  find  the  stresses  in  both 
lateral  systems  due  to  the  same  wind  pressures  as  in 
l»rob.  94. 


Art.  38.  The  Factor  of  Safety  for  a  body  is  tlie  ratio 
of  its  breaking  to  its  wDikiiij;  stress;  or  it  is  tlu'  ratio  of 
the  load  which  will  just  erusli  the  body  to  the  assumed 
load,  or  load  whieh  it  is  intended  to  earry.  Tiuis,  if  the 
tearing  unit-stress  of  an  iron  jilate  be  20  tons  and  the  work- 
ing luiit-stress  be  4  tons,  the  faetor  of  safety  will  be  5. 

The  value  of  the  factor  of  safety  is  generally  assumed  by 
the  engineer;  different  engineers  assume  different  factors  of 
safety,  depending  somewhat  upon  the  manner  of  applying 
the  loads,  the  character  of  the  structure,  and  the  nature  and 
quality  of  tlie  material.  Thus,  for  steady  loads  and  slowly 
varying  stresses,  tin  factor  of  safety  may  be  low;  but  when 
the  load  is  applied  with  shocks  and  sudden  stresses,  tlie 
faetor  ought  to  be  large.  In  a  building  the  stresses  on  the 
walls  are  steady,  and  hence  the  factor  of  safety  may  be  low. 
In  a  l)ridge  the  stresses  on  the  different  members  are  more 
or  less  varying,  and  hence  the  factor  of  safety  must  be 
higher. 


ma. 


tons. 


Ikc  ir.j  f.'ft 

r  and  litwrr 
jses  in  both 
iures  us   in 


is  ilic  ratio 
the  ratio  of 
ho  asHiinu'd 
rinis,  if  th(! 
il  the  work- 
1  be  .'). 
assumed  by 
it  factors  of 
of  apply in;^ 
nature  and 
and  slowly 
;  but  when 
tresses,  the 
?sses  on  the 
nay  be  low. 
rs  are  more 
ty  must  be 


nniDOE  TRUSSES. 


ll'J 


Also,  it  has  been  seen  (Art.  1(5)  that  the  live  load  for 
a  short  span  is  mueh  more  tiian  that  for  a  loii^'  span.  For 
thi.s  r(>ason  the  variation  of  stiess  in  pa.ssin;.;  froni  a  loaded 
to  an  unloaded  state  is  iiiuth  j,'reater  in  the  mend)ers  of  a 
short  span  than  in  those  of  a  long  one.  Conseciuently,  the 
material  in  a  short  span  .vill  suffer  what  is  termed  fathjue 
more  than  in  a  hmg  oiu'.  And  although  the  fatigue  of 
metals  is  a  subject  not  yet  well  i;nderst<io(l,  yet  it  is  dearly 
e'-';abli-<hed  that  these  sudden  change,  of  stress  in  short 
sjjans  demand  a  larger  factor  of  safety  than  those  in  long 
8i>ans. 

In  American  practieo  the  values  of  the  factor  of  safety, 
for  steiuly  and  for  varying  stresses,  are:  from  3  to  7  f<»r 
wrought  iroti,  from  rt  to  lo  for  cast  iron,  from  4  to  S  for 
steel,  from  S  to  12  for  tind)er,  and  from  lli  to  iiO  for  stono 
or  brick. 

At  times  it  may  be  necessary  to  enploy  a  much  larger 
factor  of  safety  than  either  of  these,  owing  to  local  circum- 
stances. If  the  risk  attending  failure  (such  as  loss  of  life 
or  property)  is  small,  the  factor  of  safety  may  be  sm-iU. 
But  if  the  risk  is  large,  the  factor  of  safety  must  be  large 
in  proportion.  With  a  bridge  in  perfect  conditiim,  a  very 
small  factor  would  be  sufficient.  Hut  no  bridge  is  in  per- 
fect condition;  all  rough  places,  such  as  rail  joints,  more 
or  less  open,  produce  shocks  which  cause  sinlden  stresses. 
These  stresses  cannot  be  measured.  A  very  large  allowance 
has  to  be  made  for  these  uncertainties  and  for  the  imper- 
fect state  of  nr  knowledge ;  and  therefore  there  must  be 
a  large  factor  of  safety  to  cover  all  uncertainties. 


if 


CHAPTKK   HI. 
imiDcjK  TitUKSKH  wrrii  unkqual  distuiuution 

OF  THE   LOADS. 

Art.  39.  Preliminary  Statemonfc  —  Tlio  preceding 
<'ha|)tor  hiis  troatod  (jf  HtroHHos  iirodiiccd  by  dead  loads  and 
uniformly  distrihntt'd  live  loads.  While  this  is  the  general 
method  of  treatment  for  hiijhwaij  bridges,  and  in  English 
practice  for  railway  bridges,  it  has  become  the  general 
jinictico  "f  American  engineers  to  ealcnlate  these  stresses 
lor  railway  bridges  by  one  of  the  following  three  methods: 

(1)  The  vi.se  of  a  uuiformJii  (listi'ihiUed  excasH  had  covering 
one  or  more  panels  ft)llowed  by  a  uniform  train  load  cover- 
ing the  whole  span. 

(2)  The  use  of  one  or  two  commtnUed  excess  loads  with 
a  uniform  train  load  covering  the  span. 

(3)  The  use  of  the  actutd  specijled  locomotive  tcheel  loads 
followed  by  a  uniform  train  load. 

It  is  proposed  in  this  chapter  to  .show  how  to  find  the 
maximum  stress  in  each  member  of  a  truss  by  each  of  these 
three  methods. 

Art.  40.  Method  of  Calculating  Stresses  when 
the  Uniform  Train  Load  is  preceded  by  one  or 
more  Heavy  Excess  Panel  Loads.  — This  method  of 
linding  maximum  stresses  is  sometimes  used  to  avoid  the 
laborious  practice  of  finding  the  stresses  due  to  the  actual 
locomotive  wheel  loads  to  be  described  later. 


'»^r^^mx^^^mmmu^m!ii^^£. 


niilDCE  TItUSSKS. 


\\>\ 


JUTION 

preceding 
loads  and 
:.hi'  general 
in  Knglish 
lio  general 
80  strcHyes 
niethoils : 

id  covering 
oad  cover- 

loada  with 

vheel  luadu 

0  find  the 
!h  of  these 


68  when 
r  one  or 

method  of 
avoid  the 
the  actual 


rrt)h.  98.  A  through  double  Warren  truss,  •''ig.  41),  lias 
Id  |Kinels,  eaeh  VI  feet  long  iiud  \2  fe<*t  deep;  the  deml 
load  is  KMM)  lbs.  per  foot  per  truss,  and  the  train  load  is 


1' 


»'/ 


1       .» 

r 

r 

U 

u 

y 

•  ' 

,;' 

.r 

x> 

\/ 

/\ 

\/ 

<x 

*     * 

0 

H 

iu 

li 

to' 

»' 

d' 

4'       * 

Vis.  <«o 


2000  lbs.  per  fof)t  per  truss,  preceded  by  one 'I'v^GmOtive 
piuiel  load  of  MO  tons  pee  truss:  find  the  stresses  in  all  the 
niendn-r",. 

\V'«  may  first  find  the  stresses  caused  by  the  uniform 
dead  and  train  loads,  and  then  the  stresses  caused  by  the 
excess  loconujtivo  load,  and  add  the  results;  or,  wo  may 
determine  the  maximum  stress  in  each  menibt*r  directly  by 
one  cfpuition,  as  we  have  generally  done ;  and  this  is  the 
simplest  nutthod. 

Dead  paiud  load  per  truss  =  6  tons. 

Train  panel  load  per  truss  =  lli  tcms. 

Lcconu)tivo  ])anel  load  per  truss  =  30  tons. 

Looonujtive  excess  panel  load  per  truss  =  18  tons. 
tanfl=l;  seed  =  1.414. 

For  the  maximum  chord  stresses  (except  2-A)  the  loco- 
motive must  stand  at  4,  and  the  train  panel  loads  at  all  the 
other  joints ;  for  the  maximum  stress  in  2-4  the  locomotive 
must  be  put  at  6,  as  can  easily  be  seen.*    Then, 

Max.  stress  in  2-4 

=  2  X  6  +  T»ff  X  30  +  1^(2  +  4  +  6)  =  50.4  tons. 

Max.  stress  in  4-6 

=  2  X  18  +  4  X  18  +  (VV  -  tV)18  =  122.4  tons. 

•  This  method  does  not  give  strictly  the  maximum  stresses  in  all  the 
chord  members.  While  the  error  near  the  ends  of  the  truss  '3  quite  small, 
it  iuiTeast'S  tDwaiilu  the  iiiiiidle. 


'M 


i-l: 


if 


122 


ROOFS  ASD  nniDGES. 


The  minimum  chord  stresses  aro  due  to  dead  hja^l  alone. 

Fur  tlic  niaxininni  stress  in  any  diagonal  the  live  load 
is  placed  on  tlic  right  of  a  section  cutting  that  diagonal,  as 
by  the  usual  nietliod.     Thus, 

Max.  stress  in  5-8 

=  [1  .i  X  0  +-  30  X  ^V  + 12(1  +  3  -+  r>)]  1.414  =  67.7  tons. 
Min.  stress  in  7-10 

=  [l  x()-30x  I?,]  1.414  =  0.0. 
That  is,  the  dead  load  passing  through  7-10  to  the  left 
abutui.ut  is  neutralized  l)y  the  part  of  the  locomotive  load 
passing  through  7-10  to  the  right  abutment. 

The  following  .stresses  are  found  in  a  manner  similar  to 
the  above: 

UiTER  Chord  Stresses. 


Mkmbkkk. 

1-8 

8-5 

M 

7-9 

a-ii 

Max.  ... 

-61.2 
-16.0 

-133.2 
-  39.0 

-183.0 
-  67.0 

-210.0 
-  00.0 

-2.34.0 
-   76.0 

Min 

Lower  ( 

SiioRD  Stresses 

Meubirs, 

2-4 

■i-i) 

fi-8 

s-io 

10-12 

Max 

.Mill 

+  60.4 
+  12.0 

+  122.4 

■;  30.0 

+  176.4 
+  64.0 

+  208.8 
+  60.0 

+  226.8 
+   72.0 

Diagonal 

Stresses. 

Membibs. 

i-i 

«-6 

.vs 

7-10 

9-12 

11-10' 

Max 

Min 

+  86.5 
+  21.2 

+  71.3 
+  17.0 

+  57.7 
+  8.5 

+44.1 
+  0.0 

+  32.2 
-10.3 

+  20.3 
-20.3 

Miuc.  stress  iu  1-2  =  -  61.2  tons. 


•■s»FBai^«i»E,:' 


loiitl  alone. 
10  live  loa<l 
iliagunal,  as 


=  67.7  tons. 


to  the  left 
notive  load 

r  similar  to 


»-ii 

.0 
.0 

-2.'?4.0 
-   76.0 

10-12 

8 
0 

+  226.8 
+  72.0 

12 

11-10' 

J.2 
).3 

+  20.3 
-20.3 

lilliDaK  TI{l\*iSES. 


123 


Priili.  99.    A  <lock  Tratt  truss,  Vifi.  ")(),  has  l(t  ])anels,  each 
1'-'^  feet  long  and  ll^i  feet  (h'oi);  the  dead  load  is  S(M>  lbs. 


per  foot  per  truss,  and  the  train  load  is  1000  lbs.  per  foot 
per  truss  preceded  by  two  locomotive  i)anel  loads  i.f  ;!(>  tons 
each  per  truss:  find  the  stresses  in  all  the  members. 

Consider  three  fifths  of  the  dead  load  as  ai>plied  at  the 
upper  chord,  and  two  fifths  at  the  lower  chord.* 

Dead  panel  load  ])er  truss  =  5  tons. 
Train  panel  load  per  truss  =  10  tons. 
Excess  panel  load  per  truss  =  20  tons. 

For  maximum  chord  stresses  put  locomotive  panel  loads 
at  joints  3  and  5,  and  the  train  panel  loads  at  all  the  other 
joints.     Thus, 

Max.  stress  in  1-3 

=  4^  X  15  +  20 (-j\  -f-  ^%)  =  101.5  tons  .-=  stress  in  4-6. 

Max.  stress  in  3-5 

Max.  stress  in  5-7 

=  10.V  X  15  +  f^  (17  -I-  7  -  3)  =  199.5  tons. 

For  ma.ximum  stress  in  any  diagonal  the  live  load  is 
placed  on  the  right,  by  the  usual  method. 

*  In  railroad  bridges  it  is  customary  to  take  two  tliird.s  of  the  dead  load 
as  applied  at  the  loaded  ihord ;  that  is,  the  chord  which  carries  the  live 
load,  aud  cne  third  at  the  uuloadeJ  chord. 


11' 


124 


PGOFS  A XI)  nniDGES. 


CflonD  Strkssbs, 


.MllMIIKKH. 

1  t) 

;■-*) 

.'.-7 

■  -» 

9-11 

Max 

Mill 

-101.5 
-  22.6 

-  MJ8.0 

-  40.0 

-  iim.5 

-   52.5 

-210.0 
-  00.0 

-217.6 
-  02.5 

StRK'.SES    1"   THE    DiAGON.VI.S. 


I!  ^ 


li  'i 
I'  i' 


11 


Mrviiikhh. 

1-4 

»  l> 

.)    !* 

■  10 

i»-12 

8-9 

10-1 1 

Max..  . 
Miu.  .  . 

+ 14:!.5 
f  ;!i.7 

+  118.0 
+  20.4 

+  04.0 
+   4.0 

+  71..'J 
0.0 

+  60.2 
0.0 

+  12.1 
0.0 

+  30.4 
0.0 

Max.  compression  in  the  i)().st.)=  102.0,  00.6,  81.5, 04.5,  48.6,  iJfl.O  tons. 

Proh.  100.  A  through  Howe  truss,  Fig.  32,  has  10  panels, 
oach  12  feet  lonj;  and  12  feet  deep;  the  dead  lojid  is  1000 
Ihs.  per  foot  per  truss,  and  the  train  load  is  16(i7  lbs.  per 
foot  per  truss,  preceded  by  two  locomotive  panel  loads  of 
30  tons  each  per  truss :  lind  the  stresses  in  all  the  members. 

Dead  panel  load  =  6  tons. 
Train  panel  load  —  I'Vtons. 
Excess  panel  load  -  :'V  \>os. 

Consider  one  third  of  the  dead  ii  d  ^  applied  at  the 
upper  chord. 

CiioRi)  Stresses. 


Mkmhkrh. 

2-4 

4-8 

fi-S 

8-10 

10-12 

Max 

Min 

H  100.0 
+  27.0 

+  170.0 
+  48.0 

+  210.0 
+  (!:{.0 

+  228.0 
+  72.0 

+  2.'10.0 
+  75.0 

9-11 

1.0 
).0 

-217.6 
-  02.6 

8-9 

IC-1 1 

12.1 
0.0 

+  .'50.4 
0.0 

48.5,  30.0  tons. 

IS  10  panels, 
lojod  is  1000 
I6(i7  lbs.  per 
nel  loads  of 
lie  members. 


plied  at  the 


0 

10-12 

8.0 
2.0 

+  230.0 
+  76.0 

ItltllKiE  TRUSSES. 


I2r, 


Strkh»kh 

IN    TIIK 

DlAOONALS. 

Mkmiikiu. 

.Max.    .  . 
Mill.     .  . 

2-3 

4-5 

«-7 

&-9 

1(1-11 

7-10 

«  12 

-140.0 
-  38.0 

-123.0 
-  25.6 

-97.6 
-  8.5 

-73.0 
0.0 

-50.9 
0.0 

-9.0 
0.0 

-29.7 
0.0 

Stkkssks  in  the  Verticai.b. 


Mkmiikbk. 


Mill. 
Mill. 


!?-« 


+  104.0 
+  25.0 


6-8 


+  85.0 
+  i0.0 


+07.0 
+  4.0 


!)-10 


+  60.0 
+  4.0 


II 


+37.0 
+  4.0 


I*)-oh.  101.  A  through  Pratt  truss,  Fig.  33,  has  10  panels, 
ea'-h  15  feet  long  and  In  feet  deep ;  the  dead  load  is  1200 
lbs.  per  foot  per  truss,  and  the  train  load  is  2000  lbs.  per  foot 
per  truss,  preceded  by  two  locomotive  panel  loads  of  30  tons 
each  per  truss :  find  the  stresses  in  all  the  members. 

(Consider  one  third  of  the  dead  load  as  applied  at  the 
upper  chord. 

Chord  Stresses. 


Mkmiikks. 

2-4 

4-6 

c-s 

8-10 

10-12 

g-n 

Max.  .  . 
Mill.   .  . 

+  133.5 
+  40.6 

+  133.5 
+  40.6 

+228.0 
+  72.0 

+28.3.6 
+  94.5 

+  316.0 
+  108.0 

-322.6 
-112.6 

Stresses  in 

THE    Dl 

AOONALS. 

Mrmiirks. 

2-'l 

!i-tt 

5-S 

7-10 

»-!2 

8-9 

10-11 

+29,7 
0.0 

Max 

Min 

-188.8 
-  57.3 

+  162.7 
+   40.3 

+  118.7 
+  19.1 

+87.0 
0.0 

+  67.3 
0.0 

+  4.2 
0.0 

:  ;  \ 


r 


" 


-?  • 


Ii 


ii.f 


»■ 


ii 


126 


RooF.s  AND  niiiuaKs. 


Stkkssks  in  tiik  Vkiiticals. 


Mkuiikks. 

3-1 

.vo 

7-S 

9-11) 

11-12 

-28.5 
-  3.0 

Max 

Mill 

+  30.0 
+  0.0 

-87.0 
-10.5 

-«4.6 
-  3.0 

-43.5 
-  3.0 

Proh.  102.  A  through  Whijiple  truss,  Fig.  37,  has  VJ 
panels,  each  12  feet  long  and  24  feet  deep;  the  dead  load 
per  foot  ])er  truss  is  10(M)  lbs.,  and  tht;  train  load  is2()()()  lbs. 
per  foot  per  truss,  preeeded  by  one  locomotive  panel  load  of 
30  tons  per  truss :  find  the  stresses  in  all  the  members. 

For  max.  chord  stresses  put  locomotive  panel  load  at  joint 
4  and  the  train  panel  loads  at  uU  the  other  joints.  * 

Dead  panel  load  =  6  tons. 
Train  panel  load  =  12  tons. 
Excess  panel  load  =  18  tons. 
Max.  stress  in  4-6    =  (3  x  18  -f- 1 J  x  18 )  x  .5 

=  35.3  tons. 
Max.  stress  in  G-8    =  35.3  +  2|  x  18  =  80.3  tons. 
Max.  stress  in  8-10  =  80.3  +  2  x  18  -  xV  X  18 

=  114.8  tons. 

Chord  Stresses. 


MUMBIIBfl. 

2-4 

4-0 

6-S 

8-10 

10-1 2 

12-14 

9-11 

Max.  .  .  . 
Min.  .  .  . 

0.0 
0.0 

+  35.3 
+  9.0 

+  80.3 
+  24.C 

+  111.8 
+   30.0 

+  141.8 
+   45.0 

+  158.3 
+  51.0 

-107.3 
-   54.0 

*  Put  all  tlie  dt-ad  loftd  un  the  loaded  (^liord,  unless  otberwiso  stated. 


11-12 

-28.5 
-  3.0 


37,  has  IL' 
10  (load  load 
1  i.s  2000  Iks. 
lanel  load  of 
embers, 
load  at  joint 
ts.* 


?  tons. 
:18 


-14 

9-11 

.58.:! 
51.0 

-167..S 
-   54.0 

vf\nv  stated. 


n RIDGE  TinrssKs. 


STRESSBS    in    TIIK    l)|Aflfl.V.\LS. 


127 


Mkmhkks. 

i-i 

i-fi 

3-S 

.'-10 

T-l'.' 

!l-ll 

11-12' 

13-10 

ll-s 

Max.  .  . 

Mill.    .  . 

+  78.8 
+  20.1 

+  84.8 
+  21.2 

+  71.^ 
+  l.V 

I  +.-)0.0 

l:+  5.7 

1 

+40.0 

o.r 

+3;!.o 

0.0 

+  2.3..3 
0.0 

+  12.7 
0.0 

+.3.5 
0.(1 

Sthkssk.s  in  TIIK  Vehticai.s. 

MKMHF.ItH. 

1-2 

l-J 

.Vfi 

"-S 

'•-'" 

11-12 

i;i-u 

Max.    .  . 
Min.     .  . 

-1!.^6 
-  .S.'J.O 

-50.5 
-  0.5 

-41.0 
-  4.0 

-.32.5 
0.0 

-24.0 
0.0 

-10.6 
0.0 

-9.0 
0.0 

Pmb.  103.  A  through  Whipple  truss,  Fig.  38,  has  IC. 
panels,  each  10  feet  long  and  L'O  feet  deep;  the  dead  and 
train  loads  are  1000  lbs.  and  3000  lbs.  per  foot  jier  truss, 
the  train  being  preceded  by  one  locomotive  i»auci  .  -id  of  30 
tons  per  truss :  find  *;lie  stresses  in  all  the  members. 

ArL  41.  Method  of  Calculating  Stresses  when  one 
Concentrated  Excess  Load  accompanies  a  Uniform 
Train  Load.  —  Tliis  method  is  sonu'timt'S  iisi'd,  like  tlie 
one  ill  Art.  40,  to  avoid  the  jiractice  of  finding  the  stresses 
due  to  the  locomotive  wlieel  loads.  IWit  we  h.ave  seen  tliat 
the  method  in  Aru.  40  does  not  give  strictly  the  maximum 
stresses  in  all  the  chord  members. 

Let  Fig.  51  be  a  truss  sujiporting  a  nniform  train  load 
covering  the  span,  and  a  concentrated  excess  load  P.  We 
snppose  the  excess  load  /'  to  be  the  ditference  between  tlie 
locomotive  ])anel  load  and  the  uniform  train  ]tanel  load  as 
in  Art.  40.  Then  (lie  stress  in  any  chord  member,  us  lid, 
caused  by  the  concentiated  loud  /*,  is  eipial  to  the  tending 
moment  at  c,  the  renter  of  moments  for  hil,  divided  by  the 
lever  arm  for  the  chord ;  and  hence  the  chord  stress  will  be 


H  £;^ 


128 


nnoFs  AMt  itiaixiEs. 


a  iiiaxiinuiii  wlu-ii  tlic  roiicviitratfil  loiul  /'  is  ho  phwed  as  to 
uviiko  tlin  ImmhUh.u;  immitMit  a  iiiaxiiimm.  >Jiiw,  lor  ii  single 
concentrated  load,  the   niaxiiiiinii  liendiiig  inoniciit  at  any 


point  occurs  when  the  load  is  at  that  poii.L,  for,  if  tlm  load 
be  niovtMl  to  either  side  of  the  point,  tlie  reaction  of  the 
opposite  abutment  will  be  diminished,  and  hence  the  mo- 
ment will  be  diminished. 

Therefore,  for  a  concentrated  excess  load  and  a  uniform 
train  load,  the  maxiniuni  hendinij  moment  at  amj  point,  and 
consp(juentl;i  the  maximum  chord  stress  in  anij  member,  occurs 
when  the  concuitrated  loud  is  at  the  center  of  moments  for 
that  member,  or  at  the  vertical  section  thromjh  the  center  of 
moments,  and  the  uniform  train  lo id  covnj  the  tchole  span 
(Art.  22). 

Thus,  for  the  maximum  stres  >  in  bd,  the  concentrated 
load  P  is  at  c,  the  center  of  mom  ants  for  bd;  and  so  for  any 
other  panel  in  the  lower  chord,  or  unloaded  chord  j,'t'nerally. 
For  any  panel  in  the  upper,  or  loaded  chord,  as  he,  of  a 
like  the  Warren,  the  concentrated  load  /*  acts  at  o, 
oi  at  that  end  of  tiie  panel  which  is  to  the  right  of  the  ver- 
tical section  through  6,  the  center  of  moments  for  he,  while 
the  uniform  train  load  covers  the  whole  span  (Art.  22).  For 
any  panel  in  the  loaded  chord  of  a  truss  like  the  I'rait  or 
//owe,  where  the  apexes  of  the  upper  chord  ■"■"  vertically 
above  thos(!  of  the  lower  chord,  the  concent n. ted  load 
/'  is  at  the  apex  directly  over  or  under  the  cente-  of  mo- 
ments for  that  member. 


)  pliwi'd  iia  to 
',  for  ii  siiit,'le 
inu'tit  ill  any 


^ 


)!*,  if  l\w  load 
[letioii  of  the 
eiice  the  luo- 

nd  a  nuifunn 
Mij  point,  (ind 
uember,  (tcenra 
mo  me  II  tH  for 
\  the.  venter  of 
he  tvhvle  sjiun 

concentrated 
uid  so  for  any 
ord  j^cncrally. 
d,  as  //(■,  of  a 

/*  acts  at  0, 
lit  of  the  ver- 
s  for  he,  while 
Art.  L'li).  For 
'i  the  Pralt  or 
"•■'1  vertically 
onlr:.tcd  load 
cento.-  of  uio- 


BllIDGE  TRUSSES. 


129 


For  a  single  concentrated  load,  the  niaxinuiin  positive 
shear  at  any  section  will  occur  when  the  load  is  just  to 
the  right  of  the  section ;  for  tin-  left  reaction  is  then  a 
niaxiniuiii. 

Thi'ri'fon',  for  a  cniwcntrnlcil  ovcs.s  Imul  and  a  vniform 

train  loud,  tlw  maxininni  ttlirss  in  iini/  lirair.  occnrn  irheii  the 

cnnvcntriited   hail  in   at  thi'  jhiih'I  point  iinniediatelif  on  the 

riijht  of  the  sertion,  anil  thi-  uniform  train  load  covers  the  xjian 

from  the  riijht  ahxtment  to  this  samr  jiani'l  point. 

Thus,  the  greatest  positive  shear,  and  therefore  the 
niaxinnun  stress,  in  be  or  in  hh,  occurs  when  the  con- 
centrated loiul  is  at  the  apex  c  and  the  uniform  train  load 
extends  from  this  apex  to  the  right  abutment.  Tlie 
greatest  negative  shear  for  he  or  for  hh  would  occur 
when  the  concentrated  load  is  at  tlie  apex  h  and  the  uni- 
form train  load  reaches  from  this  apex  to  the  left  aljutment. 
For  maximum  chord  stresses,  cars  both  precede  and  follow 
the  locomotive.  This  does  not  often  happen.  For  maxi- 
mum stresses  in  the  braces  the  locomotive  precedes  the 
cars.  It  follows  therefore,  that  maximum  chord  stresses 
are  of  less  frequent  occurrence  than  maximum  web  stresses, 
which  occur  for  every  passage  of  the  train. 

Proh.  104.  A  through  Howe  truss,  Fig.  32,  has  10  panels, 
each  12  feet  long  and  12  feet  deep;  the  dead  ami  train 
loads  are  1000  lbs.  and  2000  lbs.  per  foot  per  truss,  and  the 
loconu)tive  panel  load  is  30  tons  per  truss:  find  the  stresses 
in  al'  the  members. 

Consider  one  third  of  the  dead  load  as  applied  at  the 
upper  chord. 

L*'ead  panel  load     =    G  tons. 

Train  panel  load    =12  tons. 
Excess  panel  load  =  18  tons. 


ii 


I 


I 


i;jo 


HOOFS  .\M>  niniKiKs. 


i  I 


For  the  niivxinmin  strfss  in  any  cluml  ni.'nibor,  fuj  0-8, 
tin-  fxcfss  load  is  iiliK't'd  at  the  joint  H  and  the  tniiu  load 
covi'is  the  wliole  span.     Then,  (U)  of  Alt.  19, 
Max.  stress  in 

O-H  =  lO.r,  X  18  +  T^j  X  18  X  3  =  220.8  tons. 
Otherwise  by  moments,  (1)  of  Art.  11),  thns: 
Left  leaetion  =  4.J  x  18  +  ^'^  x  18  =  '.Ki.O  tons. 
The  equation  of  moments  about  the  point  7  is 

".KIO  x  ;<0  -  18(12  +  ^4)  -  stress  in  0-8  x  12  =  0. 
.-.  max.  stress  in 

(]-H  =  f),'}.0  X  ;{  -  18  X  3  =  220.8  t<ins,  as  before. 
For  the  maximum   stress   in   any  diagonal,  as   4-5,  the 
rxcoss  load  is  plaeed  at  (J,  and  the  train  loads  cover  tb.e 
sjian  from  this  point  to  the  right  abutment.     Then, 
Max.  stress  in 
4-r»  =  -  [in  X  0  +  1.2(1  +  2  +  -  +  8)  +  .8  X  18]1.4U 

=  -111.1  tons. 
Min.  stress  in 
0-7  =  -  [2i  X  0  -  I J  X  3  -  T^ff  X  18]1.414  =  -  11  tons. 


I 


CiioitK   Stokssks. 


Memiikkh. 

2-1 

4-« 

B-s 

ft-io 

10-12 

Max 

Mill 

+  07.2 
+  27.0 

+  172.8 
+    48.0 

+  22tl.8 
4-    (!;J.O 

+  250.2 
+    72.0 

+  270.0 
+    75.0 

:    :' 


i 


StBESSKS     IX    TIIK 

])lA(SONAI.8. 

Memiikkh. 

a-.-} 

i-r< 

i:-'i 

!-9 

Ki-n 

T-IO 

'J-12 

Max.    .  . 
Min.    .  . 

-130.7 
-  38.0 

-111.1 
-  25.4 

-80.5 
-11.0 

-6:5.0 
0.0 

-42.4 
0.0 

-6.1 
0.0 

-22.9 
0.0 

iibcr,  as  n-<S, 
le  tiiiiii  load 


tons. 

s. 
s 
12  =  0. 

Iwfdrc. 
as  4-5,  the 
lis  cover  the 
'hen, 

X  18]1.414 


—  11  tons. 

-10 

10-12 

59.2 

7-.'.o 

+  270.0 
+    75.0 

7-10 

5-12 

-5.1 
0.0 

-22.9 
0.0 

nniDGE  ritustiEs. 


StHKHXKH   I.N    THE    V'kKTIC'ALS. 


131 


Max. 

MCMIIIM. 

»   i 

6-6 

■-S 

S-IU 

11-12 

+  34.0 

+  4.0 

+  95.2 
+  25.0 

+  70.0 
+  10.0 

+  59.2 
+    5.8 

+43.0 
+   4.0 

Mill 

I'liih.  105.  A  tliniuijh  Pratt  truss,  Fi^-  ."Vl,  has  10  panels, 
each  ir»  feet  Ion;.;  iinil  ]'>  feet  dci'i);  the  dead  and  train  loads 
art'  IL'OO  ilis.  uiid  L'lMK)  His.  per  foot  per  truss,  and  the  excess 
loud  is  1*0  tons  per  tri;«s  :  tuid  tiie  stresses  in  ail  tiie  inenihers. 
Consider  one  third  of  tlie  dead  load  as  applied  at  the  upper 
chord. 

Dead  jiane)  load  =  9  tons. 
Train  jianel  load  =  15  ions. 
Excess  panel  load  =  20  to:is. 

Max.  stress  in  8-10  =  lOJ  x  24  -f-  f^  X  -0  X  3  =  294  tons. 
Max.  stress  in 

3-G  =  m  X  9  +  U  X  36  +  f  J  X  8]  1.414  =  143.0  tons. 
Max.  stress  in 

7-8  =  -  [1|  X  9  +  3  -f  1.5  X  21  -I-  2  X  6]  =  -  60  tons. 

Chord  Stresses. 


Mbhbkiu. 

8-» 

*s 

6-S 

8-10 

10-12 

9-11 

Max.  .  . 
Min.  .  , 

+ 120.0 
+  40.6 

+  120.0 
+  40.6 

+  224.0 
+  72.0 

+  294.0 
+  94.5 

+  330.0 
+  108.0 

-350.0 
-112,5 

Stressks 

IN    THE 

Diagonals. 

Mkmhkhh, 

Max.    . 
Mill.     . 

2-) 

c-c 

5-8 

7-10 

9-12 

t-9 

10-11 

-178.2 
-  40.5 

+  14.3.5 
+  39.0 

+  111.0 
+  19.8 

+  80.0 
0.0 

+62,3 
0.0 

+  2.1 
0.0 

+20.2 
0.0 

*4 


t 


■ym 


i^ 


■««p 


•  i! 


Vi2 


HOOFS  AS  I)  nil  I  DUES. 


t 


SrilBSHKH    IN    TIIK    VERTirALi. 

Mkmukim. 

S-t 

6-6 

7-S 

W-IO 

ll-ii 

Max. 

.Mill. 

41.0 
+  0  0 

-81.6 

-fllt.O 

-  40,0 

-  3.0 

-20.0 
-3.0 

-17.0 

-   3.0 

I'nih.  106.  A  double  Warren  truss,  Fig.  49,  used  as  a 
dock  bridge,  lias  10  pauels,  each  14  feet  loug  and  11  feet 
deep;  the  dead  and  train  loads  are  1000  lbs.  and  I'OOO  lbs. 
per  foot  per  truss,  and  tiie  excess  load  is  20  tons  per  truss: 
liud  the  stresses  in  all  the  members. 


L'pPKii  C'litniit  Strk»se.«. 


.Mkuhkkk. 

13 

it-.'i 

-154.0 
-   42.0 

.V  * 

M» 

'.MI 

Max 

-00.0 
-14.0 

-213.0 
-  O.i.O 

-2.')  1.0 
-   77.0 

-208.0 
-   84.0 

Mill 

Lower  Chord  St 

KES.<<KS. 

Memukhs. 

2-» 

4-r, 

c-s 

^-!(l 

III  U 

Max 

+  70.5 
+  17.6 

+  152.6 
+  45.6 

+  213.6 
+  00.5 

+  25.",.  5 
+   80.5 

+  272.6 
+   87.. "i 

Miu          

DiAiiONAL  Stresses. 


Mkmbekk. 

i-» 

8fl 

5-8 

7-10 

9-12 

U-lo' 

Max 

Min 

+  82.0 
+  19.8 

+00. 3 
+  10.0 

+  50.0 
+  0.3 

+  30.0 
-11.5 

+  23.2 
-23.2 

+  11.5 
-30.0 

Max.  stress  in  1-2  =  2^  x  21  +  10  =  62.5  tons. 


U-U) 

ll-ii 

-40,0 
-  3.0 

-2(i.O 
-3.0 

10,  usod  as  a 
^  and  14  feet 
ami  L'OOO  Ihs. 
oils  per  truss : 


T  U 

SMI 

i.-.I.O 
77.0 

-208.0 
-   M.O 

^-!ll 

III  12 

J53.5 
80.5 

+  272.5 
+  87.5 

»-12 

U-lo' 

+  23.2 
-23.2 

+  11.5 
-■'JC.O 

>ns. 


nitihcK  ritrssEH. 


13:{ 


rri>h.  107.  A  ilf.k  Trutt  truss.  Fig.  W),  has  10  paiu'ls, 
caili  l.-»  tVct  loiiK  and  IT*  feet  di-fp;  tlie  d.'ad  and  train  loads 
lire  SOO  ll»s.  and  L'OOO  llis.  jicr  foot  jit-r  truss,  ami  the  oxccss 
load  is  L'(»  tons  per  truss:  tind  the  stresses  in  all  the 
inend)ers. 

Art.  42.  Method  of  Calculating  Stresses  when  two 
equal  Concentrated  Excess  Loads,  placed  about  50 
feet  apart,  accompany  a  Uniform  Train  Load.  — This 

method  is  often  used,  like  the  two  in  .\rts.  In  ami  41.  to 
avoid  the  praetiet!  of  iinding  the  stresses  due  to  the  hx'O- 
motive  wheel  loads,  and  is  a  nearer  approximation  to  the 
aetual  loads  than  either  of  the  i)ther  two. 

i^et  Fig.  *■»-  lie  a  truss  siipporting  a  uniform  tr.ain  load 
eovering  the  s])an,  and  two  equ.al  eoneeiitrated  exeess  loails, 


(^„A— ^fi 


Vitf.  Oid 


/',  and  Pj.  As  in  Arts.  40  and  41,  we  suppose  ea^h  excess 
load  to  he  the  ditference  hetween  the  loeomotive  jianel  load 
and  the  iniiforni  train  panel  lo.ad.  Then  the  stress  in  any 
ehord  nieud)er,  as  bd,  eaused  by  the  concentrateil  loads  J', 
and  I\  is  eciual  to  the  bending  moment  at  <;  the  center  of 
moments  for  M,  divided  by  the  lever  arm  of  bd ;  and  hence 
the  chord  stress  will  be  a  maxinnim  when  the  two  con- 
centrated lo.ads  are  so  jHaeed  as  to  make  the  bending 
moment  a  maximum. 

Now,  for   two  ecpial   loiids,  a   tixed  distance   apart,  the 
maximum  moment  at  any  point  occurs  when  one  load  is  at 


i, 


» 


r 


:  [' 


H 

?': 


; 


184 


HOOFS  ASH  nun  (iki.fi. 


tlit>  |i(iiiit  and  tli(>  (>tlit<r  i.s  on  tlit>  Iuiik'i'I'  M>g;nont  uf  tlui 
tniHS.     'I'liiH  may  1m'  shown  iw  follows: 

Ltit /=  iiMi^'lli  t)l'  Hpan  in  Fi),'.  ol',  </ =  distiuice  between 
till'  two  t'(|uiil  loiulH  /',  anil  /'j,  and  <«  =  distance  from  left 
alnitnn'nt,  to  ifntcr  of  moments  c;  tiien  supiiosinf,'  tlic  load 
/',  tt»  1m'  at  a  distance  x  to  the  left  of  c,  an<l  eallinf,'  .1/  the 
moment  at  the  point  c,  duo  to  tho  two  loads,  /*,  and  1\,  and 
denoting  eiu-h  hnid  liy  /',  sinuo  they  are  equal,  we  have 

M  =  P -^  -'2a-d)-(l-2)x 

—  plk  "  ~  "),  a  maximuui,  when  x  =  0. 

Similarly,  if  the  load  /*,  be  placed  at  a  distance  x  to  tlm 
right  of  c,  the  moment  at  the  point  c  will  be  a  maximum 
when  X  =  0. 

Therefun',  for  tiro  (>(pial  concentrriteil  excess  loiuh,  and  a 
nni/orm  tntiii  lo<i'l,  the  vxtximum  bending  vumicid  at  ainj 
point,  anil  ninKi-i/iicHll;/  the  ni<i.riinvm  chord  stress  in  any 
Vicndicr,  occurs  when  one  of  the  eipud  concentnileil  lodls  is  nt 
the  center  of  moments  for  that  nieniher,  and  the  other  con- 
centrated load  is  on  the  lomjer  seyment  of  the  truss,  while  the 
uniform  train  load  covers  the  whole  span. 

Thus,  for  the  maximum  stress  in  hd,  tho  concentrated 
load  /*,  is  at  c,  and  the  other  load  1\  is  to  the  right;  and  so 
for  any  other  panel  in  the  lower,  or  unloaded,  chord  of  the 
left  half  of  the  truss.  For  any  panel  in  the  upper,  or 
loaded,  chord  of  the  left  half,  as  he,  of  a  truss  like  the 
Warren,  where  all  the  members  are  inclined,  the  load  /*,  acts 
at  c,  or  at  that  end  of  the  panel  which  is  to  the  right  of  the 


pnotit  uf  till) 

lUICe    l)OtW(H!ll 

iico  IrDiii  It't't 
isiiii,'  tlic  load 
•iilliiiK  M  tlu- 
'i  and  I'„  and 
wo  liavu 


when  X  =  0. 

;aiice  x  to  the 
!  u  niaxiniuni 


loiuls,  and  ii 
mnent  ut  ainj 
Hlrcss  ill  mil/ 
(I'll  IkvIh  is  at 
the  al/ii'f  riiii- 
rims,  while  the 

concentrated 
nj,'ht;  and  so 
.,  chord  of  the 
the  upper,  or 
truss  like  the 
le  load  /*  acts 
e  right  of  the 


iiHUtr.K  rnrssKK. 


liV 


vfiliiiil  section  throiij?h  /(.the  center  of  moments  for /ic,  and 
111.'  oilier  loail  /',  is  to  Ihi-  ri,i,'iil. 

For  any  I'iinel  in  tii(<  luaihil  '•hunl  ol  a  truss  liUf  the 
l',alt  or  'lli>in\  wlinr  the  upper  apexes  are  vertiialh  above 
file  ecirrespondin);  lower  ones,  the  concenlrated  loail  /',  is  at 
the  apex  directly  over  or  under  the  eenter  of  uiouieuts  for 
that  MUMuher. 

l<\)r  two  e<pial  concentrated  loa('s  /',  and  l\,  the  maximum 

positive  shear  at  any  section  (K'curs  when  hotli  loads  ar i 

the  rij^ht  of  th.>  section  and  /',  is  as  near  to  it  as  possible; 
for  the  left  reaction  is  then  a  nuuimum. 

Therefore,  far  two  eqiinl  concent  rated  exeen»  hmla  am\  a 
uniform  train  load,  the  maximum  atreas  in  any  Imire  on-nrx 
tehen  both  loads  are  on  the  riijht  of  the  sertion  anil  one  of  them 
is  at  the  panel  point  immediidelij  on  the  riijht  of  the  sertion, 
and  the  uniform  train  load  covers  the  sjnin  from  this  point  lu 
the  riijht  abutment. 

The  second  excess  load  P^  sliould  be  placed  at  a  panel 
pt.int  at  an  interval  of  about  at)  ft'ct  to  the  right  of  /'„  to 
simplify  the  computation.  Thus,  should  the  panel  length 
be  IL',  ir»,  or  18  feet,  the  distance  between  the  loads  would 
be  48,  45,  or  54  feet. 

Prob.  108.  A  through  Warren  truss,  '^ig.  53,  has  8  panels, 
each  10  feet  long  and  10  feet  deep,  its  web  members  all 


7' 


jt' 


1' 


H  10  * 


4' 


'«' 


forming  isosceles  triangles;  the  dead  ami  train  loads  are 
1(MM(  lbs.  aiul  2000  lbs.  uer  foot  i)er  truss,  and  there  are  two 


m 


136 


uo()F<  Axn  liUihCKs. 


pxcoss  loads  50  fopt  apart,  each  of  3.']  tons :  liud  tlie  stresses 
in  all  the  inenihers. 

Dead  panel  load  =    T)  tons. 

Train  panel  load  =  10  tuns. 

Each  excess  load  —  'M  tons. 

tanfl=:  \]  seed?  =  1.117. 

For  the  maxinuini  stress  in  any  chord  menil  er,  as  4-(), 

one  excess  load  is  placed  at  (>  and  the  other  one  at  4',  or  .10 

feet  to  llie  ri'^dit  of  llie  first,  and  tiie  dead  and  train  loads 

cover  the  whole  span.     Then  I'J)  of  Art.  li), 

Max.  stress  in  4-G  ^  [0.\  x  1")  +  Y  0-  +  0)3]  x  .5 

=:114.()  tons. 
Otlierwise  by  nionuMits,  (1)  of  Art.  10,  as  follows: 
Left  reaction  =  3,^  x  1")  +  33  (|  +  S)  =-■  81.375  tons. 
The  equation  of  moments  abont  the  point  3  is 

81.375  X  15  -  15  x  5  —  stress  in  4-0  y  10  =  0. 

.-.  max.  stress  in  4-G  =  114.0  tons,  as  before. 

For  the  maximum   stress  in  any  dia;,'onal,  as  3-0,  one 

excess  load  is  placed  at  <>  and  the  otlier  50  feet  to  the  right 

of  it  at  4',  and  the  train  loads  cover  the  span  fronx  G  to  the 

right  abutment.     Tiien, 

Max.  stress  in  3-G  =  [2. J  x  5  +  V  (1  4-  2  +  -  +  G) 
4-  y  (G  4-  1)  ]  1-117  =  75.5  tons  =  -  stress  in  3-4. 
Min.  stress  in  5-6  =  -  [1|  X  5  -  f  X  10  -  |  x  33]  1.117. 
=  5.0  tons. 

I'i'PEU  CiioiiD  Stkesses. 


hT 


MrMIIEKS. 

1-$ 

i]~'i 

5-7 

7-'t 

Max. 

-80.« 
-17.5 

-147.8 
-  :}o.O 

-174,4 
-  :J7.5 

-180.0 
-    40.0 

Min     . 

il  tlie  stresses 


111  or,  us  4-0, 
10  at  4',  or  HO 
1(1  train  loads 

]  X.5 

lows : 

'5  tons. 

is 

L0  =  0. 

ofore. 

,  as  w-C),  one 

t  to  the  riglit 

from  0  to  the 

...+C) 

in  3-4. 

X  33]  1.117. 


5-7 

7-1 

14.4 
37.5 

-180.0 
-    40.0 

mU^    E  TRUSSES. 


LowKH  CiioHD  Stresses. 


Diagonal  Stkesses. 


13? 


Memhkkx. 

'2-1 

4-6 

C-4 

f-U) 

Max 

+44.8 
+  8.8 

+  114.0 
+  23.8 

+  152.8 
+   33.8 

+  174.0 

+  38.8 

Mill 

.M  KM  UKHfl. 

1-2 

1-1 

8-0 

v-s 

T-io 

Max. 

Mill. 

-100.1 
-   19.5 

+  100.1 
+   10.5 

+  75.5 
+  8.0 

+52.4 
-  5.0 

+  35.2 
-19.4 

Prob.  109.  A  deck  Pratt  truss,  Fig.  50,  has  10  panels, 
each  12  feet  long  and  12  feet  deep;  the  dea«l  and  train  loads 
are  1000  ibs.  and  2000  lbs.  per  foot  per  truss,  and  then^  are 
two  excess  loads  48  feet  apart,  each  of  30  tons :  find  the 
stresses  in  all  the  members. 

Cuonn  Stresses. 


y  K.MHKKS. 

l-i 

.?-") 

f.-7 

7-9 

9-U 

Max.     .  .  . 

.Mill 

-123.0 
-  27.0 

-210.0 
-  48.0 

-270.0 
-  03.0 

-312.0 
-  72.0 

-315.0 
-   75.0 

RriiESSEs  IN  Diagonals. 


Mkmiik)- 


Max. 

Mill. 


1-1 


+ 17.^0 
+  .38.0 


+  141.7 

+   2.'!.  8 


f-S 


+  111.1 

+     7.0 


7-10 

+  82.3 
0.0 


»-li 


s-9 


+  .16.1 
0.0 


+  7.2 
0.0 


lo-ll 


+  20.7 
0.0 


fT 


188 


ItOOFS  AND  lililDGES. 


Sthessea  in  Vekticals. 


McMltEIM. 

l-i 

n-i 

6-15 

7-8 

»-l« 

U-12 

Max 

Mill 

-138.0 
-  30.0 

-123.0 
-  27.0 

-100.2 
-   16.8 

-78.0 
-  0.0 

-58.2 
-  0.0 

-48.0 
-  0.0 

Prob.  110.  A  deck  Whipple  truss,  Y\<^.  37,  has  12  panels, 
each  12  foi't  long  and  24  feet  (i('('i>;  the  (load  and  train  loads 
are  10(M>  lbs.  and  3000  lbs.  per  foot  ]»er  truss,  and  there  are 
two  excess  loiuls  48  feet  apart,  each  of  30  tons :  find  the 
stresses  in  all  the  members. 

Dead  panel  load  =  6  tons. 
Train  panel  load  =  18  tons. 
Each  excess  load  =  30  tons. 

tan  d  =  0.5;  tane'  =  l;  sec  w  =  1.117;  sec  6' =  1.414. 
Max.  stress  in  4-6 = [3 x 24  + f§ (11  + 7) J  x. 5 =58.6  tons. 
Max.  stress  in  6-8=4x24  +  2.5(10+0)1^  =  150.0  tons. 
Max.  stress  in  3-8 

r=[2x6  +  1.5x 25+2.5x14]  1.414=119.5  tons. 
Min.  stress  in  5-10=[9-1.5x2-2.5x2]  1.414=1.4  tons. 


Chord  Stresses. 

ft, 
i 

Mp.MBKIIli. 

'i-\ 

4-0 

c-> 

t-lO 

ll!-l'i 

I'J-14 

»-il 

1 
1 

n 

Max.     . 
Mil).     . 

^no.o 

0.0 

+  .'•.8.6 
+   0.0 

+  166.0 
+  24.0 

+211.6 

+   30.0 

+286.0 
+   46.0 

+  316.6 
+  61.0 

-326.0 
-  54.0 

SA^ 


rnKm. 


9-10 

11-12 

-58.2 

-48.0 

-  0.0 

-  0.0 

as  12  panels, 

id  train  loads 

uid  there  are 

)ns:  t 

ind  the 

MM 


W^j 


6'  =  1.414. 

•=58.6  tons. 
15G.0  tons. 

tons. 

H4  =  i.4  tons. 


I '.'-14 

K-ll 

-310.6 
-  61.0 

-320.0 
-  54.0 

niiWGK  TRUSSES. 


Stiiksses  in  Diaoonals. 


139 


.Mkmuekh. 

I-l 

1-8 

i!-l 

.vio 

7-1-2 

9-14 

11-1-2' 

13-10 

ll--< 

Max 

MIn 

+  130.7 
+  -20. 1 

+  141.4 

+  •l\.i 

+  119..'! 

+  11. !i 

+97.a 

+   1.4 

+77.S 
0.0 

+.\s.n 

0.(1 

+40.8 

o.u 

+W.6 
0.0 

+  l(l.ti 
0.0 

Stresses  i.n  Verticals. 


Mkmiikks. 

1-2 

3-1 

.V3 

7-S 

9-10 

11-1-2 

lH-14 

Max.  .  . 
Min.  .  . 

-170.0 
-  30.0 

-117.0 
-  18.0 

-100.0 
-  15.0 

-84.6 
-  8.0 

-00.0 
-  0.0 

-56.0 
~-  0.0 

-64.0 
-  0.0 

Pmh.  111.  A  through  parabolic  bowstring  truss,  Fig.  44, 
has  8  jxuicl.s,  eacii  10  feet  long,  and  10  feet  center  depth; 
the  verticals  are  ties,  and  the  diagonals  are  struts;  the  dead 
and  train  loadj  are  1000  lbs.  and  3000  lbs.  per  foot  per  truss, 
and  there  are  two  excess  loads  .50  feet  apart,  each  of  30  tons : 
tiud  the  !  tresses  in  all  the  ineinbers. 

Dead  panel  load  =    5  tons. 

Train  panel  load  =  15  tons. 

Each  excess  load  =  30  tons. 

Length  of  3-4  =:  4.375  feet ;  length  of  5-0  =  7.5  feet. 

Length  of  7-8  =  9.375  feet;  length  of  0-10  =  10.0  feet. 

Then,  for  the  inaxiinuin  chord  stresses,  a  full  dead  and 
train  panel  load  must  be  at  each  lower  apex.  For  any 
ineinber  of  the  lower  chord,  as  L*-4,  one  excess  load  of  30 
tons  is  placed  at  4  and  the  other  at  (V,  or  50  feet  to  the  right 
of  the  fir.it,  and  the  dead  and  train  loads  cover  the  whole 
span,  as  just  stated.  Here  we  cannot  use  the  "method  of 
chord  iiici-eiuents,"  since  the  chords  are  not  j)arallel,  but 
must  use  the  "method  of  moments,"  (1)  of  .Vrt.  I'J. 


if 


; 


m 


140 


ROOFS  AND  nillDGES. 


m 


Reju'tion  at  the  left  end  =  3^  x  20  +  30  ( J  +  g)  -  103.75 
tons. 

Mux.  stie.ss  in  2-4  =  ^^^'^•"'"'  ^  '*'  =  237.1  tons. 

4.375 

This  tensile  stress  in  2-4  is  eijual  to  the  horizontal  com- 
ponent of  the  eoiapressive  stress  in  2-t3,  by  (1)  of  Art.  5. 
Therefore  the  stress  in  2-3  is  ecjual  to  the  stress  in  2-4 
nuiltii)lie(l  by  the  .secant  of  the  angle  between  2-^5  and  2-4. 

Thns, 

Max.  stress  in  2-3  =  -  237.1  x  1.092  ^  -  25S.9  tons. 

For  the  maximuni  stress  in  3-5,  one  excess  load  is  pnt 
at  0  and  the  other  at  4',  while  the  deail  and  train  loads 
(iover  the  whole  span ;  then  take  center  of  moments  at 
intersection  of  4-5  and  4-()  or  at  4.  And  so  on  for  the 
stresses  in  5-7  and  7-',). 

The  diagonal  .stresses  are  fonnd  by  jmtting  only  the  train 
and  excess  loads  on  the  truss  in  the  proper  position  for  each 
mend)er,  since  the  dead  load  jiriMluces  no  stress  in  the  diaj;?- 
oiials  (Art.  34).  Thius,  to  find  the  nmximiim  stress  in  4-5, 
one  excess  load  is  placed  at  0  and  the  other  50  feet  to  the 
right  of  it  at  4',  and  the  train  loads  cover  the  span  from  0  to 
the  right  abutment. 

Now  cut  3-5,  4-5,  and  4-6,  and  take  center  of  moments 
at  the  intersection  of  3-5  and  4-0,  which  is  4  feet  to 
the  left  of  2;  the  diagonal  3-0  is  not  in  aetion  for  this 
loading. 

Lever  arm  of  4-5  =  14  sin  0  =  14  x  -^p  =  8.4  feet. 

Left  reaction  for  this  loading 

=  ^•'*  (1  +  2  +  •■•+(;)  -f  ■  V  (G  +-  1)  =  05.025  tons. 

s  <s 


max.  stress  in  4-5 


(;5.0L'5  X  4 
8.4 


=  —  31.25  tons. 


+  I)  =  lOa.75 


irizontal  com- 

;i)  of  Art.  r». 

stress  ill  2— t 

2-,'5  ami  2-4. 

).S.9  tons. 

3  load  is  put 

il  train  loads 

inonu'iits  at 

\o  on  for  the 

jnly  the  train 
ition  for  each 
iS  in  the  dia^- 
^tress  in  4-.'», 
j"»()  feet  to  the 
pan  from  0  to 

•  of  momenls 

is  4   feet  to 

tion  for  this 

.4  feet. 


25  tons. 
25  tons. 


11 II I  DUE  TRUSSES. 


Ufi-KK  Ciioiti)  Stiiksses. 


141 


.Memiirrh. 

H  !1 

8-5 

&7 

7-9 

Max 

Min 

-258.8 
-   43.0 

-2;i0.r, 
-   41.0 

-213.7 
-   40.7 

-208.3 
-  40.1 

LowKu  CiiiiiiK  Sthk.hsks. 


MKMItKltS. 

■.'-I 

4-C, 

t:-^ 

^  HP 

+  220.0 
+  40.0 

Miix 

+237.1 
+  40.0 

+2.30.0 
+   40.0 

+  220.0 
+  40.0 

Mill 

StHKSSKS    in    TIIK    I)i,\ti()NAI.>'. 


.Mkmiikkh. 

4-.') 

C-T 

s-i» 

3-0 

'.-t 

7-111 

Max 

Mill 

-31.3 
0.0 

-34.2 
0.0 

-38.2 
0.0 

-40.0 
0.0 

-43.75 
0.0 

-41.1 
0.0 

Max.  stresses  in  the  verticals  =  -|-  50.0  tons. 
Min.  stresses  in  the  verticals  =  -|-  5  tons. 

Prah.  112.  \  throut,di  circular  bowstring  truss,  Fig.  45, 
has  0  panels,  each  15  feet  long,  its  web  members  all  form- 
ing isosceles  triangles,  with  the  upper  apexes  on  the  cir- 
cumference of  a  circle  whose  radius  is  75  feet,  the  center 
depth  of  the  truss  being  14|  feet;  the  dead  and  train  loads 
arc  720  lbs.  and  217«)  lbs.  per  foot  per  triiss,  and  there  are 
two  excess  loads  45  feet  apart,  each  of  24  tons:  fiud  the 
stresses  in  all  the  members. 

Dead  panel  load  =  5.4  tons. 

Train  pa«el  load  =  1G.32  tons. 

Kach  excess  load  —  24  tons. 


if 


; 


■MM 


BMM 


142 


HOOFS  AM)   li  lit  DUES. 


Ciioiti)  Stresses.  . 


Mkmbeks. 

2-8 

8-6 

5-7 

7-T 

2  4 

4  6 

6-8 

Max.  .  .  . 
Mill.  .  .  . 

-140.0 
-  24.7 

- 103.0 
-  27.2 

- 140.2 
-  25.0 

-137.1 
-  25.0 

+ 125.0 
+   20.0 

+ 130.8 
+  22.0 

+  125.4 
+  23.7 

Wur.    STIIE8.SES. 


Mkmiieks. 

ii-4 

&  « 

"-■> 

7'-«' 

5'  4' 

Max 

Mir     

+  20.0 
+  4.9 

+  33.4 
-11.0 

+  34.8 
-15.1 

+  37.2 
-13.0 

+43.2 
-  7.1 

J'rob.  113.  A  tlirout^h  Vratt  truss,  ¥\g.  33,  has  10  panels, 
eatih  12  feet  lon^  and  12  feet  deep;  tlie  dead  and  train  loads 
ar«  1000  lbs.  and  3000  lbs.  j>er  foot  per  truss,  and  there  are 
two  excess  loads  4S  feet  apart,  each  of  30  tons:  find  the 
stresses  in  all  the  uiendjers. 


Art.  13.  The  Baltimore  Truss.  — This  truss.  Fig.  54, 
or  some  modification  of  it,  is  now  used  very  generally  for 
long  si)ans  when  it  is  desired  to  avoid  lont;  panel  lengths. 
Each  large  pane'  is  divided  into  two  smaller  ones  by  insert- 


Vig.  04 

ing  half  ties  and  subvertical  struts,  or  half  stmts  and  sub- 
vertical  ties,  according  as  it  is  a  deck  or  a  tiirough  truss. 
In  the  deck  form,  Fig.  54,  the  s»il)verticals  ai,  hj,  ck,  etc., 
are  strained  only  by  the  panel  loads  which  they  directly 


4  6 

6-8 

+  130.8 
+   22.0 

+  125.4 
+  23.7 

-«' 

5'  4' 

V.2 
13.0 

+43.2 
-  7.1 

hivs  10  pan  el  .s, 

,iul  train  loads 

and  there  are 

:ous:  find  the 


triLss,  Fig.  54, 

generally  for 

[)anel  leni,'tli,s. 

jnes  hv  insert- 


* 

t'   h 

l' 

/ 

\  / 

M 

1 

V" 

trnts  and  sub- 

Ihrnngli  tr'.iKs. 

ai,  hj,  ck,  etc., 

they  directly 


noOF  TllUSSES. 


113 


siipjinrt;  the  dotted  diagonals  4k,  CI,  G'm,  4'n  are  counter.s, 
not  licing  in  acition  foj;  full  load. 

Proh.  114.  .\  deck  Haltiniore  truss,  Fig.  /i4,  ha.s  10  panels, 
each  10  feet  long  and  20  feet  deep;  all  the  verticals  are 
po.sts,  an<l  all  the  inclined  pieces  are  tics;  the  dead  and 
train  loads  are  ,S()0  Ib.s.  and  1000  lbs.  per  foot  per  truss,  and 
there  are  two  excess  loads  rtO  feet  apart,  each  of  30  tons : 
find  the  stres&es  in  all  the  members. 

Dead  panel  load  =  4  tons. 
Train  panel  load  =  8  tons. 
Each  excess  load  =  30  tons. 

The  maximnm  chord  stresses  occur  when  the  dead  and 
train  loads  act  at  every  upper  apex  (Art.  20),  and  the  excess 
loads  act  at  tlu;  proper  apexes  (Art.  41i).  Thus  for  .'{/*.  we 
have  ,"0  tons  at  b  and  at  9,  50  feet  to  the  right  of  h.  The 
center  of  moments  is  at  4,  the  intersection  of  3-1  and  2-4. 
ilence,  the  left  reaction  for  dead  and  live  loads  is, 

iJ  =  7^  X  12  -I-  f « (13  4-  8)=  129.375  tons. 
Calling  S  the  stress  in  3  ft,  we  have 

129.375  X  40  -  12  (30  -(-  20)  +  A'  x  20  =  0 ; 

.•.  S  =  stress  in  3  ft  =  —  228.8  tons  =  stress  in  65. 

The  stresses  in  1  a  and  a  .'i,  3  ft  and  ft 5,  5 c  and  c7,7 d  ami 
f?9,  etc.,  must  always  be  the  same,  since  the  jjosts  ai,  hj,  rk. 
etc.,  are  perpendicular  to  the  chords,  and  cannot  therefore 
cause  any  stresses  in  them. 

For  5  c  and  (-7.  we  uni.st  put  the  excess  loads  at  c  and  7'. 
and  take  the  center  of  moments  at  G,  the  intersection  of  5  A 
and  4-6;  and  so  on. 


"If' 

I 


II 


I     . 


(J  t 


Mil  : 


p 


144 


HOOFS  AND  ItltlDGES. 


r. 


For  any  lower  chonl  nit'inlu'r,  as  -Mi,  we  havo  'M)  tons  at 
n  and  at  e,  50  feet  to  tlie  ri;,'ht  of  T).  The  center  of  moments 
is  at  6.     Thus, 

/i!  =  7J  X  12  +  ^«  (lli  +  7)=  lL'r).r.2r)  tons. 

.-.  125.025  X  40  -  12  (.SO  +  20  +  lO)-  .S  x  20  =  0. 

.-.  S  =  stress  in  4-0  =  215..'{  tons. 

It  is  evident  that  the  niaximnm  stress  in  each  subvertieal, 
(i(,  hj,  ck,  ill,  ete.,  is  vi]nid  to  a  full  panel  load,  4  +  8  +  30 
=  42  tons  conijjression. 

It  is  also  evident  that  half  of  the  load  on  ai,  bj,  (Jf,  etc.,  is 
transmitted  to  .'{,  5,  7,  etc.,  through  the  hal;  ticj  ».'i,  j5, 
k  7,  etc. 

Therefore  the  maxinunu  tension  in  each  ol  the  half  ties 

jn,j5,  A;7,etc.,  =  '*-±4-'t-^secd  =  21  x  1.414  =29.7  tons.' 

The  nuixinuun  stress  in  the  upper  part  of  any  main  tie, 
as  ok,  (H'ciirs  when  the  excess  loads  act  at  c  and  at  7',  50 
feet  to  the  right  of  c,  and  the  train  loads  are  at  all  the 
apexes  on  the  right  of  5  k.  The  shear  in  5  A-  for  this  load- 
ing is, 

3^x4  +  fg(ll  +  6)  +  ^«j(ll  +  10+-  +  l)=78.875. 

.•.  max.  stress  in  5A;  =  78.875  x  1.414  =  111.5  tons. 

The  maximum  stress  in  any  main  vertical,  as  5-4,  is 
equal  to  the  greatest  shear  in  5-4 ;  and  this  greatest  shear 

1  At  the  center  of  the  truss  these  ties  must  .ict  .is  countem.    Thus,  take 

the  tie  Wl,  put  tlie  excess  lo.ids  at  3  and  il,  and  tlio  train  loads  at  all  the 

ape-xes  from  the  loft  abutment  to  il  inclusive.    Then  the  greatest  shear 

in  the  section  cutting  (Ml ,  Mt,  ami  l~H  is  —  I'S.ST.'i  tons.    As  /-8  is  not  in 

action  for  this  loailin;^,  this  is  the  vertical  component  of  the  stress"  iu 

i-U. 

..max.  stress  in  l-'J  =  ^8.875  X  1.414  =  +  40.8  tons. 


niHhCE  THVSSES. 


ijr> 


ve  'iVS  tons  at 
or  of  uioiiit'iits 

IS. 

X  *-'0  =  0. 


ch  subvertical, 
id,  4  +  8  +  30 

\,  bj,  (k,  etc.,  is 
h'  tie  J  ».'3,  j5, 

'.  the  half  ties 

4  =29.7  tons.' 

any  main  tie, 
(•  and  at  7',  r»0 
are  at  all  the 
k  for  this  loud- 

1)=  78.875. 

1.5  tons. 

3al,  as  5-4,  is 
greatest  shear 

Hter!>.  Thus,  take 
ill  loads  at  all  the 
the  sireate.st  slioar 
.  As  /-8  is  not  in 
t  uf  the  stres!>  iu 

ituns. 


in  5-1  occurs  wlien  the  excess  loads  act  at  5  and  at  c,  fiO 
fict  l<>  tlie  ri^,'ht  of  5,  and  the  train  loads  are  at  all  the 
apexes  from  the  ri^'ht  abutment  to  5  inclusive.  Then  the 
greatest  shear  in  5-4  is  that  which  is  due  to  this  loading, 
together  with  the  four  dead  loads  at  5,  c,  7,  and  (/,  half  of 
tlu!  dead  load  at '.),  and  half  of  the  dead  load  at  b,  which  is 
transmitted  to  5  through  the  half  tie  j5.     Hence, 

Max.  stress  in  5-4  = 

-[5  X  4  +  fo(l-  +  ")+  A(12  +  -  +  1)] 
=  -  94.0  tons. 

The  maximnm  stress  in  the  lower  part  of  any  main  tie, 
as  ./I,  is  equal  to  the  maximum  stress  in  4-5  intdtiplied 
by  the  secant  of  the  angle  which  the  tie  makes  with  the 
vertical.     Hence, 

Max.  stress  in  >  4  =  94.6  X  1.414  =  133.8  tons. 

Tlie  maximum  stress  in  6-Z  occurs  when  the  excess  loads 
are  at  a  and  7,  and  the  train  loads  are  at  all  the  joints  from 
the  left  abutment  to  7  inclusive.  Then  the  shear  in  the 
section  cutting  rf-9,  /-9,  and  Z-8  is  —  21.G25  tons,  which  is 
the  vertical  comp(ment  of  the  stress  in  M),  since  l-H  is  not 
in  action.  The  load  at  d  produces  in  /-9  a  negative  shear 
of  2  tons,  so  that  the  difference,  or  19.625,  must  come  from 
the  member  6-^ 

.-.  max.  stress  in  C>-1  =  +  19.625  x  1.414  =-.  +  27.7  tons. 

Otherwise  as  follows:  The  live  loads  going  to  the  right 
abutment  through  the  panel  6-8  =  23.63  tons.  The  dead 
loads  crossing  this  at  the  section  cutting  f/-9,  /-9,  and  l-H  are 
I  of  load  at  9  +  .[  of  load  at  d  -  4  tons.  The  difference  is 
19.()3  tons,  for  which  the  panel  6-8  must  be  counterbraced. 

Thus  are  found  the  following  maximum  stresses : 


1M5 


HOOFS  AMt  iiiniiCKS. 


Maximi^ 

Clloltli   SrllKHWKH. 

1  il 

ii-'i 

6-7 

7  9 

ii-i 

4-9 

+  201.1 

-130.0 

-228.8 

-281.0 

-205.0 

+  127.1 

f2l.-..;5 

M.v.xiMiM  SniKHHKs  IN  riM'KU  I'akt  OF  Main  Tik)*  ani>  IIai.1-  Tiks. 


1  t 
+  11(3.5 

H 

b-k 

;-; 

•A-i 

6-J 

7* 

4-40.8 

+  151.1 

+  111.6 

+  74.H 

+  20.7 

+  20.7 

+20.7 

.Mavimim  Stiikhsks  in  LowKii  I'.Mti  <>i    .Main  Tiks. 


i  i 

./-» 

k  i\ 

;-H 

(>^i 

+  0.7 

+  174.8 

+  133.8 

+  05.0 

+  00.3 

+27.7 

Maximum  Sthe'^sks  in  tiik  VKnTicAi.s. 


ft'^^i 


i-A 

4-5 

B  7 

S-9 

«/,  h},  ck,  (I!. 

-123.0 

-94.0 

-07.0 

-59.0 

-42.0 

Note.  —  In  the  above  determination  of  the  niaxiinum  Ktresses  in 
ttie  long  verticals  and  in  the  lower  ends  ',  f  the  long  diauon.-ils,  a.s,  for 
exiiniiilc,  in  5-4  and  j-4,  it  was  assumed  thiit  the  larjrest  p(>s.sibk'  shear 
in  5-4  or  in  J-4  occurs  when  the  excess  loads  act  at  5  and  at  f,  and 
the  train  loads  are  at  all  the  apexes  from  the  rij;ht  ahutinent  to  the 
joint  5  inclusive.  It  is  evident,  however,  from  ;i  little  inspection  of 
KIg.  54,  that,  if  a  train  pane!  load  be  placed  at  llie  joint  /(,  one  half  of 
this  train  load  at  b  will  be  transndlted  to  the  joint  5  through  the  half 
tie  j5,  just  as  one  half  of  the  dead  load  at  6  is  tri-nsuiitled  too  through 
the  half  tie  j5,  and  that  this  half  train  load  transmitted  to  5,  minus 


'S^, 


*««:.,,. 


•1-0 

<^s 

21. -..;'. 

+  2(11.1 

AND    II.M.I-   TiKH. 


T-* 

ii-i 
4-40.8 

+20.7 

N    'I'lKS. 


4-t 

.7 

+n.7 

(II,  fij,  ck;  ,1!. 


-42.0 


imuiu  Htresses  in 
ilingoiials,  as,  for 
est  possible  sliear 
u  ami  at  f,  and 
,  abutment  to  tlio 
;tle  inspection  of 
inl  h,  om;  half  of 
tliroiifrli  tlif  liiilf 
tti'd  toi")  tliroiifili 
litted  to  5,  ininiis 


nitllXlK  THUSSES. 


11" 


tlip  part  that  is  tran.iinittcd  to  tlic  ri«lit  abninuni,  wliicli  is  ,'^  of  tin- 
train  load  at  h,  will  form  part  of  tlit>  shear  in  5-4  or  in  j-\  ;  thai  is, 
the  shear  will  l)u  iiicrciuti'il  hy  J  x  8  —  ,9g  x  H  =  ,"4  x  8  =  2  5.     Ik'nec, 

Max.  strfsaln  &-»  =  -  [5  x4  +  JX(12  +  7)  +  .'4(l2  +  •••  +  lj+2.5] 
=  -07.1  tons. 

Max.  8tn>88  in  j-\  -  1)7.1  x  1.11 »  =  l.'H.:)  tons. 

Similarly, 

Max.  Blri'HH  in  2-^  =  -  [7  x  12  f  };i(  U  )  !•)  ]  -      127. 1  tons. 

Max.  Htrt'HS  in  i-2  -  127.1  x  I.  II I  --  1711.7  tons, 
and  HO  on  for  U-7  and  k-i\,  8-1),  and  /-8. 

Thus,  stn.ss  in  «-7  -  -  (i8.(5,  in  A-fl  -  07.0,  in  8-!»  =  -  4.1.1,  in 
/-8  -  (11  tons. 

It  is  cvidunt  that  by  putting;  n  train  panel  load  on  the  apex  just  to 
the  left  of  the  section  considered,  the  shear  in  that  section  is  inere-i-sed 
the  most  when  near  the  left  end  of  the  triisH  ;  and  that  it  liecreases  on 
approaching  thu  ndddic  of  the  tru88  without  beuondng  zero. 

l',"'i.  115.  A  deck  liultiinort;  truss,  Fi^'.  .W,  has  14  panels, 
each  10  feet  loiij.,'  aiul  L'(»  feet  (leep,  the  verticals  being  posts 
and  the  diagonals  ties ;  the  dead  and  train  loads  are  1(M)0  lbs. 

7      a     3      h     ,'!     e     7     a      7'   e     J'    f      fa      V 


w 

\/ 

\/ 

\l/ 

H^ 

\/ 

\/ 

\/m 

\ 

\ 

/\ 

A 

A\ 

/ 

/ 

a  «' 


4' 


and  2000  lbs.  ])er  foot  per  trnss,  and  there  are  two  excess 
loads  50  feet  apart,  each  of  30  tons:  find  the  inaxiniiiin 
stresses  in  all  the  members.     (See  note  to  last  problem.) 

Maximum  C-iord  Stuf.sses.  - 


l-s 

;!-5 

r» -7 

7  7' 

•i-i 

4-fi 

+  214.:i 

t;-i'.' 
+  260.7 

-142.6 

-2.30.4 

-271.1 

-204.7 

+  130.7 

•«.'?•' 


I. 


' 


iii' 


MH 


hoot's  A.M)  iiiiiiiaHs. 


Maximum  Sthknhki*  in  Upi-ku  Part  <»k  Main  'I'ikh  ani>  IIai.k  Tik.*. 


+  201.6 

»-J 

6-* 

7-V 

+ni.H 

+:ji.8 

7-* 
+  31.8 

+  ltlO,0 

+ 102.6 

+  00.1 

Maximi  M  Stuksskn  in  Lowi.h  I'aht  or  Main  Tn.n. 


l-i 

J-i 

i~n 

W 

4-* 

+  184.8 

+  v.n.n 

+  85.8 

+  42.4 

+  8.1 

Maximi'M  SmKOfiKH  IN   riiK  Vkuticals. 


^» 

4-.'i 

ft-T 

,ii,  hj.  ,1,  ,11. 

-i:jo.7 

-MM 

-00.7 

-4:).o 

/'/•(»/>.  116.  A  through  Biiltimore  truss,  Fig.  WJ,  has  Ifi 
jianels,  eacli  Hi  feet  h)ng  and  .'51-'  feet  (Um'Ji  ;  the  dead  load 
is  given  liy  formula  (.'{),  Art.  ir»,  tin;  train  load  is  luUO  lbs. 


.t  fi 


7'  X-  S' 


JUT  foot  per  trus'-.,  and  there  are  two  excess  loads  4S  feet 
apart,  each  of  .'><•  tuns-  find  the  inaxiinuni  .stresses  in  all 
the  nienibcrs. 

Head  panel  load    -  (^Ar^:^~!^^^  =  «  tons. 


1,'  X  L'(HM) 
Train  iiancl  load    —  12  tons. 
Each  excess  U)ad  --  oO  tons. 


HS^, 


■teS, 


ANi>  IIai.i--  'I'ik*. 


b-J 

+  31.8 

h:n.8 

UN    TlK*. 

4-* 

+  8.1 

L». 

((,  /,J.  iX,  '//. 

-4.').0 

'ijj.  Tifi,  hiis   1(5 

the  (load  load 

•ad  is  luUO  Ib-s. 


s' 


u     k'    h    ^  t' 


s  loads  4S  feet 
.stresses  in  all 


^  8  tons. 


nit  1 1)111-:  iiirssics. 


ll'.» 


Tlit^  niaxinuini  cliord  stiPs.srs  iK-ciir  when  tlio  dead  and 
train  loads  att  at  every  lower  apex  (Art.  '_'(>),  and  the  excess 
loads  act  at  the  proper  apexes  (,\rt.  I-).  Tims  for  /><>,  \vc 
have  :W  tons  at  I  iind  at  c,  •«  feet  to  the  ri^'lit  of  (>.  The 
center  of  nionients  is  at  .'$,  the  iiiterseetion  of  li~."»  and  .'{-0. 
Hen(!0  the  left  reactiuM  for  dead  and  live  loads  is, 
i?  =  7i  X  liO  +  ,'!!  (I  t  -f  11)  =  190.875  ton.s. 

Calling  s  the  max.  stress  in  iO,  we  have, 

I0().87r.  X  .'!•-'  -  20  X  IT)  4-  20  x  U\  -  s  x  32  =  0. 

.•.  s  =  max.  stress  in  hO  =  llMi.l)  tons  =  max.  stress  in  46. 

For  any  upper  chord  meinhcr,  as  5-7,  we  have  ,".0  tons  at 
8  and  at  c,  IH  feel  to  the  right  of  8.  The  center  of  moments 
is  at  8.    Thus, 

ii  =  7,\  X  20  +  ^H  (10  4  7)  =  181.87.-.  tons. 
.-.  181.875  xOx  16-20(1(5  +  32 +  48  +  (U-f-80)  +  .sx32=0. 
.•.  8  =  max.  stress  in  5-7  =  -  3;)5.(»  tons. 

Similarly,  the  other  chord  stresses  are  found. 

It  is  evident  that  the  m,'<.xinnim  stress  in  each  snhvertical, 
ai,  bk,  d,  dm,  etc.,  is  etpial  to  a  full  panel  load  8  -|-  12  +  .'>0 
=  50  tons  tension. 

It  is  also  evi<lent  that  half  of  the  load  on  (d,  hk,  rl,  dm, 
etc.,  is  transmitted  to  4,  4,  (5,  8,  etc.,  through  the  half  stmts 
1 4,  A:  4,  '(5,  )»8,  etc. 

Therefore  the  maximum  compression  in  each  of  the  half 
struts  a,  A- 4,  /G,  viS,  etc. 

=  8  +  12  +  3()  ^p^.  Q  ^  2r>  x  1.414  =  .35.4  tons, 
o 

The  maximum  stress  in  the  lower  part  of  any  main  tie, 
as  /8,  occurs  wlien  the  excess  loads  act  at  8  and  at  f,  48  feet 
to  the  riglit  of  8,  and  the  train  loads  are  at  all  the  apexes 


"?   ■ 


i 


i  "' 


i  I 


f 


k 


%{ 


150 


HOOFS  AND  liRlDGES. 


from  the  right  abutment  to  8  inclusive.     The  shear  in  /8  lor 
this  loading  is, 

2i  X  8  +  -Jg  (10  +  7)  +  H (10  +  ...  4- 1)  =  03.125. 

.-.  max.  stress  in  18  =  93.125  x  1.414  =  131.7  tons. 

The  maximum  stress  in  the  upper  part  of  any  main  tie, 
as  3 A;,  occurs  when  tlio  excess  loads  act  at  G  and  at  (/,  48 
feet  to  the  riglit  of  G,  am  tliu  train  loads  are  at  all  the 
apexes  from  the  right  abutment  to  b  inclusive :  for  half  of 
the  train  load  at  b  is  transmitted  to  the  joint  G  through  the 
half  strut  kG,  and  f^^r  of  the  train  load  at  b  is  transmitted  ft) 
the  right  abutment;  therefore  the  difference  of  these,  or  ^\ 
of  the  train  load  at  b,  goes  to  the  left  abutment  and  forms 
part  of  the  shear  in  3/i';  that  is,  the  train  panel  load  at  b 
will  increase  the  sliear  in  '6k  by  yV  X  12  =  3|  tons,  it  is 
evident  also  that  half  of  the  dead  load  at  b,  which  is  trans- 
mitted to  G  through  the  half  stmt  kC>,  goes  to  the  left  abut- 
ment and  forms  part  of  the  shear  in  3k.  Similarly  for  5/ 
and  7  m.     Hence, 

Max.  stress  in  3  A; 

=  [">  X  8  4-  fga-  +  9)  +  H  (12  +  -  + 1)  +  3.75]  1.414 
=  200.3  tons. 

The  maximum  stress  in  3-4  is  ecjual  to  the  full  panel  load 
at  4  plus  half  the  sum  of  the  panel  loads  at  a  and  b 

=  ,30  -j-  12  -I-  8  -f  I  (24  +  IG)  =  70  tons. 
The  maxiuium  stress  in  either  of  the  other  long  verticals, 
as  7-8,  is  equal  to  the  shear  in  7  m  plus  whatever  load  is 
applied  to  the  upper  apex  7.     Hence, 
Max.  stress  7-8 

=  -[8  +  M(8  +  «''>)+1|(8  +  -  +  1)+tVx12] 
."=-60.1  tons. 


ns», 


iliear  in  /S  lor 

=  93.125. 

.31.7  tons. 

any  main  tie, 
!  and  at  (/,  48 
ire  at  all  the 
; :  for  half  of 
G  throut,'h  the 
transmitted  to 
)f  these,  or  j^^r 
!nt  and  forms 
mel  load  at  /* 
I  tons.  It  is 
,liich  is  tran.s- 
the  left  abut- 
niilarly  for  5/ 


+  3.75]  1.414 

full  panel  load 

and  b 

)ns. 

long  verticals, 

atever  load  is 


rV  X  12] 


JililhGK  TRUSSES. 


mi 


The  maximum  stress  in  3  ("  occurs  when  the  excess  loads 
act  at  4  and  at  c,  and  the  train  loads  cover  the  whole  tr  i 's  ; 
it  is  easily  se«;n  that  lialf  of  the  dead  train  loads  .(  (■, 
which  are  transmitted  to  4,  form  part  of  the  shear  r  3;'. 
Hence, 

Max.  stress  in  3 1  =  -  [7  x  20  +  f  f  (14  +  11)]  1.414 
=  --  2W.2  tons. 
The  following  stresses  are  found  in  a  manner  similar  to 
the  above : 

Maximum  Ciiokd  Stuesses. 


S  6 

r,-7 

T-!) 

2-» 

4-6 

o-s 

8-10 

-318.8 

-395.6 

-417.6 

+  100.9 

-106.9 

+  328.8 

+405.0 

Maximum  Stresses  in  Lower  Paiit  of  Main  and  Half  I/iaconals. 


i-i 

A--8 

/-3 

HI-IO 

(-4 

i-4 

1-6 

m  S 

-283.7 

+  189.3 

+  131.7 

+  78.3 

-35.4 

-3>.4 

-35.4 

-35.4 

Maxi.mcm  Stresses  in  Upper  Part  of  Main  Diagonals. 


8-i 

3-* 

6-1 

i-m 

9-m 

t-l 

-204.2 

+200.3 

+  140.5 

+  85.0 

+  53.2 

+6.2 

Maximum  Strf-sses  tn  the  Verticals. 


S-4 

5-« 

T-S 

8-10 

ai,  bk,  ol,  dm. 

+  70.0 

-90.4 

-00.1 

-23.9 

+  50.0 

IS   i 
I   f. 


il  ■ 


w 


152 


HOOFS  AND  nillDOES. 


The  stress  in  0-)ii  •-=  9-u  occurs  whoa  the  excess  loads 
act  at  »'  1111(1  iV,  and  the  train  loads  act  at  all  the  joints 
from  the  riyht  ahulnicnt  to  c.  'rhen  the  shear  in  the  sec- 
tion cutting  y-7',  '■)-((,  and  !(»-/(  ^^  lOJ.OL'o-S  x  .S=  -{-o'-iVJo 
tons;  and  as  lO-ii  cannot  take  compression,  this  is  the 
vertical  component  of  stress  in  9-)i. 

.-.  max.  stress  in  9-u  =  -f  37.025  x  1.414  =  +  53.2  tons. 

If  10-u  were  constructed  to  take  compression,  this  would 
ho  greatly  modified.  In  this  case  the  excess  loads  would 
act  at  8'  and  ij,  and  the  train  loails  ut  all  the  joints  from  the 
right  abutment  to  8'.  Then  the  left  reaction  =  92.G25  ton.s 
and  the  shear  in  the  section  cutting  l)-7',  9-)i,  and  10-jt  = 
9'_'.(525  -  CA  =  -+-  28.025  tons.  The  member  10->i  takes  a 
positive  shear  of  4  tons,  so  that  the  vertical  component  of 
stress  in  9-?i  =  +  28.(>25  -  4  =  +  24.025  tons. 

.-.  max.  stress  in  \)-n  =  4-  24.025  x  1.414  =  +  34.8  tons. 

The  stress  in  7'-0  occurs  when  the  excess  loads  act  at 
f  and  4',  and  the  train  loads  at  all  the  joints  from  the 
right  abutment  to/.  Then  the  left  reaction  =  -f  "^4.375  tons 
and  the  shear  in  the  section  cutting  7-5',  7-0,  and  8-0  = 
81.375-8  x  10=  -f  4.375  tons.  As  8'-0  does  not  act  in 
compression,  this  shear  must  be  resisted  by  ^'-O  alone. 

.-.  max.  stress  in  7'-0  =  +  4.375  x  1.414  =  +  0.2  tons* 

Prob.  117.  A  deck  Baltimore  truss.  Fig.  54,  has  10  panels, 
each  10  feet  long  and  20  feet  tleep,  the  verticals  being  post^ 
and  the  diagonals  ties;  the  dead  and  train  loads  are  1000 
lbs.  and  2000  lbs.  per  foot  per  truss,  and  there  are  two 
excess  loads  50  feet  apart,  each  of  33  tons :  find  the  stresses 
in  all  the  members. 

•It  should  be  noted  that  the  members  m-8  and  ;i-8'  are  also  tension 
members  for  couiiipr  stresses. 

The  teusiuu  stress  iu  each  of  these  members  =  20.025  x  1.414  :=  29.2  tons. 


mi 


Jumt 


excess  loads 
ill  the  joiiiLs 
ir  ill  tlic  scc- 
;S=  +;>7.G2u 
,  this  is   the 

+  53.2  tons. 
»n,  this  would 
i  loads  would 
lints  from  the 
=  92.G25  tons 
,  and  10-jt  = 
10-?i  takes  a 
component  of 

+  34.8  tons, 
loads  act  at 
nts  from  the 
+  "^A.Tib  tons 
0,  and  8'-0  = 
33  not  act  in 
-0  alone. 
+  6.2  tons* 

lias  IG  panels, 
Is  being  post^ 
)ads  are  1000 
here  are  two 
d  the  stresses 

are  also  tension 
1.414  =  29.2  tons. 


nUIDGE  TRUSSES. 


Maximum  CnoKD  Stbesses. 


15:} 


l-l 

S-,'> 

.  ''  ' 

T  it 

2-1 

4  6 

fi  ^ 

-104.1 

-274.1 

-337.7 

-354.7 

+  152.4 

+258.1 

+  317.8 

Maximum  Stresses  in  Uppeii  Pakt  of  Main  and  Half  Ties. 


l-i 

»-J 

r,  i- 

:-/ 

37 

+34.0 

.'■'-; 

--<: 

+47.4 

+  232.0 

+  180.0 

+  132.7 

+  88.3 

+  34.0 

+  34.0 

Maximum  Stuesses  in  Lower  Paht  of  Main  Ties. 


1-2 

J~i 

1-6 

/-S 

6-/ 

+  215.6 

+  164.1 

+  116.2 

+  09.1' 

+  31.8 

The  counter  4-k  is  found  not  to  be  required;  but  in  actual 
practice  a  counter  would  bo  used  to  provide  for  loads  differ- 
ing from  the  ones  above  assumed. 

Maximum  Stresses  in  the  Verticals. 


2-3 

4-5 

6-7 

S-9 

fii,  hi,  ei;  ,11. 

-152.4 

-110.1 

-82.2 

-09.2 

+  48.0 

This  problem  is  taken  from  "Strains  in  Framed  Strnctnres"  by 
A.  J.  Du  Bois.  It  will  Ik-  seen  that  the  answers  above  given  for  the 
maximum  stresses  in  the  lower  part  of  the  nuiin  ties  and  in  the  long 
verticals  differ  soinrwhat  from  those  given  by  Prof.  Du  Rois. 


"n 


f  ' 


h 


n 


154 


ItOOFS  AND  BllIDGES. 


Art.  44.  True  Maximum  Shears  for  Uniform  Live 

Load.—  Tliiis  iar,  it  h;is  been  assmiu'd  Uiat  the  live  panel 
loads  were  all  coiici'iitiated  at  the  panel  points,  and  in  order 
to  find  the  niaxiniiun  stress  in  any  diagonal  dne  to  a  nniforni 
live  load  per  linear  foot,  the  niaxinuini  live  load  shear  in 
that  diagcnial  has  been  found  by  putting  a  live  panel  load 
on  every  joint  on  the  right  of  the  seetion  cutting  the  diagonal 
(Art,  22). 

For  exaniide,  it  has  been  assumed  that  the  first  apex  on 
the  right  of  the  section  in  Fig.  57  has  a  full  live  ])anel  load 
aiul  lilt'  first  one  on  the  left  has  no  live  load.  Now  the  joint 
n  on  the  right  of  the  section  receives  the  half  panel  load 


A7\KA7gg\,| 


t„        ;■'". — >M J 

Kip.  07 

between  the  joints  n  and  n  - 1,  but  only  a  small  part  of  the 
load  on  the  left  of  n ;  the  other  part  of  the  load  on  the  left 
of  n  is  received  by  the  joint  n  + 1.  When  the  uniform  load 
extends  from  the  right  abutment  to  the  joint  n  +  1,  the  joint 
})  receives  a  full  panel  load ;  but  then  the  joint  n-\-\  receives 
a  half  panel  load.  Let  us  therefore  ascertain  at  what  dis- 
tance X  to  the  left  of  the  joint  ?i  the  load  must  extend  to 
produce  the  greatest  shear  in  the  n  +  1th  panel. 

I  A>t  N=  the  number  of  panels  in  the  truss,  p  =  the  panel 
length,  w  =  the  uniform  live  load  i)er  linear  foot,  n  =  the 
number  cf  whole  panels  covered  by  the  load,  li  =  the  reac- 
tion at  the  left  or  unloaded  end.     Then, 


:  the  part  of  the  load  ivx  which  is  carried  by  the  joint 

n'+l;  and  wx  —  ^  =  the  part 
2p 

carried  by  the  joint  n. 


of  the  load  wx  which  is 


rnif  orm  Live 

the  live  panel 
;s,  ami  in  order 
le  to  a  uniform 

load  shear  in 
ive  panel  load 
ig  the  diagonal 

3  first  apex  on 

live  i)anel  load 

Now  the  joint 

lalf  panel  load 


lall  part  of  the 
oad  on  the  left 
le  uniform  load 
n  +  1,  the  joint 
it  n  +  l  receives 
in  at  what  dis- 
nust  extend  to 
lel. 

i,  p  =  the  panel 
ir  foot,  n  =  the 
i,  It  =  the  reac- 

ied  by  the  joint 
id  wx  which  is 


ItlllDflK  TliUssES. 
We  have  then  for  the  reaction 

li  ^'-je:^"  + ') 


'jp 

wxn 

'"IT 


+ 


IV):' 


wn^p 


'2pN-  •    '2N 
Hence  the  shear  in  the  n  +  1th  panel  is 


~~N'^2pX 


'■""      2p 


2N 


155 


\  2p)N      2N 


(1) 


(2) 


Equating  the  derivative  of  Fto  zero,  we  have, 


N^pX~J~' 


X  = 


n 


N-1 


P, 


which  is  the  distance  to  the  left  of  tlie  joint  n  that  the  live 
load  must  extend  in  order  to  produce  the  niaximuni  siiear 
in  the  n  +  1th  panel  from  the  right  end. 

Thus,  let  the  truss  have  8  panels;  then  for  the  true 
maximum  shear  in  the  sixth  panel  from  the  right  end 
X  =  ^p,  for  the  seventh  panel  x  =  S-p,  and  for  tlie  eighth  or 
last  panel  x=lp  =  p,  or  tlie  whole  truss  is  covered. 

The  total  live  load  on  the  truss 

=  mcp+J^^  =  ^!lIEP 
N-1     N-1 

=  ^  times  the  load  on  the  n  +  1th  panel. 

Thm'forp,  the  lirp  had  on  the  n  +  1th  pvicl  As  ^  th  of  the 
total  lire  load  on  the  triins.  '^ 

Substituting  the  above  value  of  x  in  (2)  and  reducing,  we 

Maximum  shear  =  — ^-^^ 

2  (N-1) 


ft  i 

i  I 


t 


ir,»! 


HOOFS   AM)   lUllhClEfi. 


These  valiips  of  a*  and  the  inaxinmin  sliear  are  iii(U'j>eii(lent 
of  the  form  of  the  tniss^  wlifthrr  llie  weliliiiij,'  be  iiicliiietl, 
as  in  llie  Warren,  or  inelined  and  vtMliciil.  as  in  tlie  Tratt 
and  Howe. 

Pi-oh.  118.  A  trns.s,  Kit,'.  r>7,  has  7  jiaiiels,  eaeh  10  feet 
lonK',  tlie  live  h)ad  is  2  tons  per  foot:  iind  (1)  the  true 
niaxinnim  live  load  shears  in  every  i)anel  and  (2)  the 
niaxinmm  live  load  shears  by  the  usual  method  (Art.  22). 

Here  N=l,p  =  10  feet,  and  to  =  2  tons. 
For  the  maximum  shear  in  the  1st  panel  from  the  left 
n  —  (5. 

...  .T  =  ;.  =  IC.  feet,  and  shear  =  ^ili^l*'=:9(J  tons, 
'  2  X  (> 

which  is  half  of  the  effective  load,  or  the  effective  reaction. 

For  max.  shear  in  the  4lh  ])anel  from  left,  n  —  'A. 

.•.  x=  I  X  !(■)=:  S  feet,  or  the  load  reaches  to  the  middle 

of  the  4th  panel. 

8hear  =  g.xj<>  ^ '^'=  24.0  tons. 
2  X  (> 

By  the  usual  way  of  taking  a  full  panel  load  as  con- 
centrated at  the  third  apex  from  the  right  n  -  1,  and  none 
at  the  4th  apex  n,  we  have 

Shear  =  -^i  (1  +  2  +  3)  =  27.4  tons, 
which  is  greater  than  that  obtained  by  the  strictly  correct 
method. 

Thus  the  following  live  load  shears  are  computed  (1)  by 
the  Strictly  Correct  Method  and  (2)  hy  the  Method  in  Common 
Practice : 

True  Method.  90.0,  00.7,  42.7,  24.0,  10.7,  2.7,  0.0  tons. 
Common  Method.  90.0,  08.0,  4r>.7.  27.4.  1.3.7,  4.0,  0.0  tons. 


-<tj.if 


ro  iiiiU>i)Oii(lent 
\\i  lie  iiicliiu'ti, 
s  in  tlic  Vvixil 

i,  oiuili  K)  I'eet 
I  (1)  the  tine 
1  and  (2)  the 
od  (Art.  22). 

tons. 

from  the  left 


H2 

-  =:9()  tons, 


ective  reaction. 

'ft,  n  r=  3. 

i  to  the  middle 


il  load  as  con- 
i  —  1,  and  none 

ns, 

strictly  correct 

!omputed  (1)  by 
hod  in  Common 

7,  2.7,  0.0  tons. 
7,  4.C,  0.0  tons. 


•PWI 


iinihuK  ritussKs. 


157 


U  is  seen  that  Hit'  .sliciirs  dlilaiiit'il  liy  tlie  ?(.s»(((/  nn'tlioil, 
that  is,  liy  siipitosinji;  the  [lani'l  loads  to  Ixs  coneentrati'il  at 
the  panel  |)oints,  are  larger  than  tlio.se  obtained  hy  the  true 
mclliiiil.  For  liie,  reason  that  the  shi'ais  obtained  by  the 
\isnal  method  iihvays  err  on  the  safe  side,  and  that  the  error 
is  always  small,  it  is  the  common  practice  to  snppose  all  the 
load  from  middle  to  middle  of  panel  to  be  concentrated  at 
the  panel  point,  as  in  Art.  4. 

Art.  45.  Locomotive  Wheel  Loads.  —  In  compntin^ 
the  stresses  in  railway  bridges  in  America,  the  live  load 
which  is  now  generally  taken,  consists  of  two  of  the  heaviest 
engines  in  nse  on  the  line,  at  the  head  of  the  heaviest 
known  train  load.  The  weights  of  the  engines  and  tenders 
are  assumed  to  1)0  concentrated  at  the  wheel-bearings, 
giving  definite  loads  at  these  {)oints,  while  the  train  load 
is  taken  as  a  uniform  load  of  about  3000  pounds  per  linear 
foot  of  single  track. 

(a.) 

^    nonho    noP.o    ^    hnnnn    o  o  o  o{^^>!l 

(b.)  (ItO/eel  =  il  inch  tcole) 

Fig.  58. 

The  first  diagram  of  Fig.  r)8  represents  two  88-ton  pas- 
senger locomotives,  as  specified  by  the  Pennsylvania  Rail- 
road. 

The  second  diagram  shows  two  112-ton  decapod  engines, 
used  on  the  Atlantic  Coast  Line. 


\    \ 


(  t 


I  i 


m 


158 


HOOFS   AM)   nilUHiKS. 


The  tiiinibers  aliovc  tli«'  wImmOs  sliow  their  weights  in  tons 
for  hotli  rails  of  a  .siii.L;lc  triu-k.  Tlie  numbers  between  the 
wlieels  show  their  distaiiees  apart  in  feet. 

Art.  46.  Position  of  Wheel  Loads  for  Maximum 
Shear. —  Let  Fi};.  iV.)  represent  a  truss  with  a  (leea|io(i 
engine  and  train  upon  it. 

Let  N=  the  number  of  panels  in  the  truss,  ]i  =  tlie  ])anel 
length,  7'= /',  4  /'a +•••+/'«=  the  sum  of  the  wheel 
loads,  ?«  =  the  uniforju  train  load  per  linear  foot,  U'— the 
total  live  load  on  the  truss,  d  =  the  distanee  of  the  center  of 


gravity  of  /*  from  the  front  of  the  train,  x  =  the  distanee  of 
the  front  of  the  train  from  the  right  abutment,  a  —  the  dis- 
tanee of  /',  from  the  front  of  the  train,  y  =  its  distance  at 
the  left  of  the  nth  panel  point,  and  li  =  the  reaction  at  the 

left  end.     Then  J\  •'  =  the  part  of  the  load  ]\  that  is  carried 

P 
by  the  n  -f  1th  panel  jjoint. 

We  have  then  for  the  reaction 


iV>  '>Np 

Hence  the  shear  in  the  u  4  lili  panel  is 

r    — -r-z 1-  —  I  I     ■ 

Up  2  Nu  p 


(1) 


(^) 


n 


litithGE  TIlUSSES. 


15!) 


figlits  in  toiiH 
i  betwei;a  the 


'  Maximum 

Lli  a  (li'ca|io(l 

p  z=  tlip  ])aiu'i 
t'  the  wIh'I'I 
fdot,  W=  the 
'  the  center  of 


k 


16  distance  of 
b,  u  =  tlie  (lis- 
ts distance  at 
Raction  at  the 

that  is  carried 


(1) 


(-',) 


But  from  the  fifjure,  x  +  «  =  tip  ■\-  y ; 
which  in  (-')  gives 


y  ^  (I  -f  X  -  np, 


Np       •  2Np      /"  +  ^-"^')- 
Equating  the  first  derivative  of  Fto  zero,  we  have, 

P  +  1VX 


X 


-y'.  =  0; 


or,  since  P  +  ivx=  W,  we  have, 

Ilencr,  the  shear  in  an>i  panel  ix  a  maximum  when  the  load 
on  the  imnel  is  —th  of  the  entire  live  load  on  the  truss. 

In  practice  it  is  convenient  to  put  one  of  the  U)ads  at  tlie 
nth  panel  pijint,  so  that  the  above  condition  cannot,  in  gen- 
eral, be  exactly  satisfied.     We  must  have  in  general, 

P,  =  or<lTr. 

Hence,  in  general,  the  shear  in  the  n  +  1th  panel  is  a 
maxiniitm  ivhen  one  of  the  loads  is  at  the  nth  jianel  point,  and 

the  load  on  the  u  +  1th  panel  is  equal  to  or  just  less  than  — Ih 


of  the  entire  load  on  the  timss} 


N 


Prob.  119.  A  through  Warren  truss  has  10  panels,  each 
12  feet  long :  find  the  maximnm  shears  in  each  panel  caused 
by  a  single  decapod  engine  and  tender. 

Here  each  driver  in  Fig.  T)!)  weighs  12.H  tons,  each  tender 
wheel  weighs  10  tons,  and  the  ]tilot  wheel  weighs  8  tons; 
the  total  load  IF  is  112  tons. 

1  This  holds  good  in  all  casus  (or  shear. 


I     I 


i  * 


u^>AiWA0MMMKmv>rMMi« 


iT 


H\ 


1«',0 


II OOFS   ANFt  UlUltnKfi 


fict  it  !>(<  n'i|ini'i'i|  In  timl  tin-  iiiaxiiiuiin  slicar  in  the  2d 
piuifl  tn.iii  tlic  Icll.     llciK  «  ==  S;  S      1<». 

Hy  the  aliovi'  rule  for  tlio  iniixiimnn  sliciir  in  this  iniiicl, 
the  first  driver  must  W  put  ;it  llif  joint  h,  since  H  <  '1',^-,  and 

In  tliis  position  of  the  loads,  tlie  left  reaction  is 
/;=  A.  X  1(»4  +  -f^  C.»(;.(»  f  91 .75  +  87.r.  -f  m.'l't  +■  70.0) 

4  JiL(71.5  +  GG.83  +  C1.23  +  5G.5f))=74.0  tons. 

/.  max.  .shear  =  74.0 -y-  =  G9.C  tons. 

Let  it  1)0  rcpiired  to  find  the  niaxinuini  shear  in  the  8th 
panel  from  tlie  left.     J  lore  7*  =  1.'. 

To  iind  the  jHisition  of  the  wheels  for  maximum  shear  in 
this  jianel,  try  iho  first  driver  at  the  joint  u,  as  lieforo. 

For  this  position  the  total  live  load  on  tlie  truss 

=  8  +  5  X  12.8  =  72  tons, 

since   the  tender  wheels  are  oft'  the  truss.     T\v  the  rule, 
therefore,  this  is  not  the  correctt  position,  since  8  >  j'g  x  72. 
If  we  put  the  pilot  wheel  at  71,  the  load  on  the  truss 

=  8  +  4  X  12.8  =  50.2  tons, 

since  the  last  driver  is  off  the  truss.     Hence  for  maximum 
shear  in  this  panel  this  is  the  correct  luadini?. 

With  the  live  load  in  this  position,  the  left  reaction  is 

iJ  =  A  X  24  +  ^|:y  IG  +  11.75  +  7.5(»  +  3.25)=  5.7  tons, 

which  is  also  the  maximum  shear,  since  in  this  case  there  is 
110  load  to  he  suhtraeted. 

Thus  the  fullowin};  shears  are  determined: 

Max.  shears  =  80.8,  GO.G,  58.4,  47.2,  3G.0,  24.8,  13.0,  5.7. 


T^'''^«r**:s  ^^x,"*? 


liciir  ill  the  '2d 

•  ill  tlii.s  luiufl, 
le  8  <  ',7,  aud 

ion  is 

t.25  +  70.0) 

50) =74.'.)  tons. 

liear  in  the  8th 

(iiiiiini  shear  in 
as  lii'l'ore. 
truss 

Tiy  tlie  rule, 
the  truss 

5  for  maximum 

r 

left  reaction  is 
25)=  5.7  tons, 
lis  case  there  is 

{4.8,  13.9,  5.7. 


nillDGE   llii'.'SES. 


tni 


Pfoh.  120,  A  deck  I'ratt  truss,  Fij,'.  50,  has  10  jiani'ls, 
each  20  feet  long  and  24  feet  deep:  find  the  inaxiiuuiu  web 
str(^sst's  in  e;u^h  panel  caused  by  a  sinnU;  pas.senger  locomo- 
tive and  tendt-r,  as  shown  in  {(()  of  Fij,'.  58. 

\\i)  tiist  lind  tiie  niaximnin  shear  in  the  left  ])anel.  I>y 
the  i-\\\{'  the  second  pilot  wheel  must  be  put  at  the  joint  4, 
since  H<  jp„  and  8  -f  8>  flJ. 

With  the  live  load  in  this  position,  the  left  reaction  is 

Ji  = ,  ^  (185.5  4-  I'^O  +  IG.'J.r*  +  148.5  +  14.S  -f- 138) 
+  11^-^(171  + 163)=  71.34  tons; 


aud  the  max.  shear  =  71.34  —  8  x 


5J) 
20 


:  09.14. 


.-.stress  in  l-l  =  09.14  sec  6  =  09.14  x  1.302  =  90.0  tons. 

For  3-t  we  have  the  same  loading  as  for  1-4;  and  the 

maximum  compression  in  3-4   is  equal   to  the  shear  just 

found  for  1-4. 

.-.  stress  in  3-4  =  —  09.1  tons. 

The  maximum  compression  for  the  end  vertical  1-2  is 
found  by  placing  the  wheels  so  as  to  bring  the  greatest  load 
to  the  joint  1,  and  is  found  to  be  71.9  tons. 

Thus  we  find  the  following  maximum  stresses: 

Diagonals  = -I- 90.0,  +78.0,  +07.1,  +55.0,  +44.2,  +.32.7, 
+  21.2,  +  9.8  tons. 

Posts  =  -71.9,  -09.1,  -00.3,  -51.5,  -42.7,  -33.9  tons. 

If  the  second  driver  be  placed  at  the  middle  panel  point  II,  the 
Ktress  in  11-12  will  be  found  to  be  -  39.0  tona. 

Prob.  121.  A  deck  I'ratt  truss,  Fig.  .50.  has  10  panehs, 
each  12  feet  long  and  12  feet  deep ;  the  dead  load  is  1000 


;->«tftfc*»'ia<T-jir... 


1 


162 


llOoFf^   AM'   ItUlhUKS. 


llis.  \M'v  liiu'iir  fiiDt  iif  track,  tlic  livr  load  is  a  i>a,ss(MiK'(>r 
lociniKitivo  ami  tt'inlcv,  as  hIiowii  in  (n)  of  Kig.  M:  timl  the 
iiiaxiiiiuin  web  stresses  in  oach  paiu'l. 

Tho  (Icatl  and  live  load  stresses  may  lie  comimted  sepa- 
rately. 

HtRKHHK*    in     DudliNM.S. 


Mkmiikkh. 

1-4 

H-ll 

&-■« 

7-l() 

+  12.7 
+  61.(1 

+03.7 

ii-li 

11-10' 

!''->' 

Dead  strpftHcs. 

+  3H.0 
+   HH.3 

+  21t.fl 
+  76.0 

+  106.6 

+  21.1 
+03.4 

+  H4.6 

+   4.2 

+  3H.6 

+  42.7 

-  4.2 
+-20.1 

-12.7 

+-13.7 

Max.  BtresaoH . 

+  120.;! 

+-21.9 

+-   1.0 

Kthksskm  in  Tiir.   I'owts. 


MfMltKliS. 

1-.' 

.-. 

.',-« 

T-^ 

!l-ll> 

U-l'J 

Dead  htrcswcs . 
Live  KtrosHPS  . 

-  30,0 

-  70.0 

-27.0 
-02.6 

-21.0 
-63.7 

-16.0 
-44.0 

-  9.0 
-30.1 

-  fl.O 

-27.:i 

Max.  Htroiwes  . 

-100.0 

-80,6 

-74.7 

-69.9 

-46.1 

-33.3 

If  tlie  first  (liiviT  !»•  jiUvcd  at  llii'  iiiiililli'  imihI  point  11,  tlio  stress 
in  11-12  will  lie  fimiul  |.)  be  -  2H,7  tens,  I;  will  In-  noticiMl  tliat  the 
tension  of  4.2  tons  in  tin-  iliasonal  0'-12  is  iMiiiivaicnt  to  a  c'oinpre.ssioii 
of  4.2  tons  in  the  diagonal  11-10' ;  and  siinilavly  for  the  next  diagonal 
0'-8'. 

Proh.  122.  A  through  Pratt  truss,  Fig.  GO,  has  8  panels, 
each  18  feet  long  and  24  feet  deep :  find  the  niaxinuun  web 


; 3' 


*  Ji^  ,        0,;  ,..* 


'-■-§1- 


imiiKiE  mvstiKs. 


108 


,s  a  i>a.sH(Mijjpr 
[.  r*« :  timl  i\w 

oinpiued  sepa- 


11-10 

!''->' 

2 

7 

-  4.2 
+20.1 

+21.9 

-12.7 

+  i;i.7 

+   1.0 

(1-1(1 

U-l'i 

-  9.0 

-30.1 

-  0.0 
-27.:) 

-45.1 

—  M>..» 

)iiit  11,  tlio  stress 

noticud  tliiit  the 

to  a  c'oinim'ssioii 

the  next  diagonal 


),  has  8  panels, 
maximum  web 


X.. 


Hlifsscs  ill  each  iiaiicl  due  to  a  |titss('ii^,'t'r  locoimitivc  and 
It'iidiT,  as  .shown  in  (/()  of  Kit;,  aS.  followed  by  a  uniform 
train  load  of  .'UXK)  lbs.  per  linear  foot. 

(1)  To  find  the  niaximuni  stress  in  2S. 

Try  the  first  driver  at  the  panel  })oiiit4.  In  this  position 
the  total  live  loa<l  on  the  truss 

==«S  +  '.M>  X  l.r.  =  22.3  tons. 

Sinee    S+.S<23\   and    S  +  8  +  20>JJ^   tliis    is   the 
eorrect  position  for  the  maximum  shear  in  the  panel  2-4. 
The  reaction  is 

/f  =  -^^  aaS  +  140.6  -f- 108.5  +  10.'i.r>  -l-  0»  +  93) 
144 

+  ,tt(^-'''  +  11«)+  ?.  X  -T^,~  =  113.77  tons. 
144  Z  144 

.-.  max.  shear  =  113.77  -  ~-  (0  -f-  14.5)=  103.33. 

18 

.-.  stress  in  2-3  =  -  103.33  x  1.25  =  -  129.2  tons. 

(2)  To  find  the  maximum  stress  in  9-8'. 

Try  the  first  driver  at  the  joint  8'.  In  this  position  the 
total  live  load  on  the  truss 

■    =88  +  18  X  1.5  =  115  tons. 

But  8-f  8>  Lp;  hence  this  is  not  the  correct  position 
for  the  maximum  shear  in  the  panel  10-8'. 

We  therefore  try  the  second  pilot  wheel  at  8'.  We  then 
have,  for  the  total  live  load  on  the  truss, 

Tr=  88  +  9  X  1.5  =  101.5  tons. 

Since  8  <  — — ->  this  is  the  correct  position  for  the  maxi- 
o 

mum  shear  in  the  panel  10-8'. 


f.-i?t«i««*vi*M«»«/F*!iT.»»«T^Mm:-'  T3ii^---it  i^vnxikfi'i 


Im 
Ml 


n 


m 


lot 


HOOFS  AND  DllIDGES. 


The  reaction  is 

1540  4-  IfilO  4-  13.r>  X  4.5 


11  = 


=  22.51  tons. 


144 

.-.  max.  shear  =  22.51  -  ?-^-i'-''^  =  20.1  tons. 

18 

=  —  stress  ill  9-10. 
.-.  stress  in  9-8'  ::=  20.1  x  1.25  =  25.1  tons. 

The   max.  tension  in  the  hip  vertical  3—1   is  fouinl  by 
puttiiii,'  tlie  wlioels  so  as  to  bring  the  greatest  load  at  4. 
Tims  iho  following  uiaximum  stresses  are  found: 
Diagonals 

=  -129.2,  +9G.4,  +07.9,  +43.6,  +25.1,  +11.3  tons. 
Verticals  =  +  36.9,  -  54.8,  ~  34.9,  -  20.1  tons. 

Art.  47.  Position  of  Wheel  Loads  for  Maximum 
Moment  at  Joint  in  Loaded  Chord.— In  addition  to 
lh(!  notation  of  Art.  4(!,  let  P'  —  the  load  on  the  left  of  the 
panel  point  n,  Fig.  59,  x'  =  the  distance  of  its  center  of 
gravity  from  the  point  n,  and  n'  =  the  number  of  panels 
between  the  left  abntuient  and  the  point  n.  Then  for  the 
moment  at  the  panel  point  7i,  Fig.  59,  wc  have 


■P(a;  +  (r)       K'j;- 
Xp  li~Xj>_ 


n'p  -  Fx',  [by  (1)  of  Art.  46]. 


Equating  the  first  derivative  of  M  to  zero,  we  have,  since 

N  ' 

or  since  P+  ivx—  TF(Art.  40),  we  have 


N 


""■I 


niilDGE  TliUSSES. 


165 


1  tons, 
tons. 


;ons. 

-4  is  found  l)y 
itpst  load  at  l. 
found : 

5.1,  +11.3  tons, 
tons. 

or  Maximum 

-  In  addition  to 
1  tlio  left  of  the 
if  its  center  of 
mhcr  of  panel.s 
Then  for  the 
ve 


(1)  of  Art.  4G]. 
,  \vc  have,  since 


Jlence  the  moment  nt  (tni/  paiui  pniitt  in  the  hailed  chord  is 

v' 
a  maximxm  when  the  UkuI  on  the  lejl  of  the  point  is    -ths  of 

the  entire  lice  lo(((l  on  the  triiH.^. 

In  practice  it  is  convenient  to  put  one  of  the  loads  at  the 
Hti.  panel  point,  so  that  the  above  condition  can  very  seldom 
l)e  exactly  satisfied.     We  must  have  in  general 


P' 


or  <  -  W. 

N 


Hence,  in  general,  the  moment  at  any  panel  point  of  the 
loaded  chord  in  a  maximuni  ifhen  the  load  on  the  left,  of  the 
point  has  to  the  entire  load  on  the  trusn  a  ratio  which  is  equal 
to  or  J II xt  less  than  the  ratio  irhich  the  nnmher  of  panels  on 
the  left  of  the  point  bears  to  the  entire  number  of  panels  in  the 
truss. 

By  tliis  rule  the  maximum  chord  stress  in  any  member  of 
the  upli)adcd  clii  rd  may  be  determined,  since  we  have  only 
to  divide  the  maximum  moment  at  the  panel  point  opposite 
the  chord  mendier  by  the  depth  of  truss. 

Prob.  123.  A  deck  Pratt  truss,  like  Fig.  50,  lias  8  panels, 
each  18  feet  long  and  24  feet  deep:  find  the  maximum 
chord  stresses  due  to  a  single  passenger  locomotive  and 
tc'uder. 

Let  it  be  required  to  find  the  maximum  stress  in  the 
second  panel  3-5.     Here  n' =  2,  X=S,  ll'=S,S;  therefore, 
by  the  rule,  J"  must  be  equal  to  or  less  than  5  x  88,  or  22 
tons.     Hence,  the  lir,st  driver  must  be  put  at  5  since 
8  +  8  <  22,  and  8  +  8  +  20  >  22. 

With  the  live  load  in  this  position,  the  left  reaction  is 

H  ^   ^-  (122.5  +  117  4-  90.5  +  85.5  +  80  +  76) 


144 


+  -'i  (108  -I-  100)  =  G0.58  tons. 
144  ^ 


'-^,  -.tim/itMttt/'M  i----.^. 


1G(; 


ROOFS  AND  li  HI  DOES. 


.-.stress  in  3-:>  =  ~l^^^^^^^^^^^^^±^^  =  -  83  ton. 

=  —  stress  in  C-8. 
Thus  aro  foinul  the  following  stresses: 
Max.  stress  in  l-.'?:^  —47.7,  in  3-5=  -83,  in  5-7=  —103.8, 
in  7-".)  =  -  110.5  tons. 

I'i'cli.  124.  A  (li'ck  I'ratt  truss,  Fig.  50,  has  10  iiancls, 
each  20  feei  hnvj;  and  L'4  feet  deep:  find  llic  nia.\iniiini 
chord  stiesses  in  each  panel  due  to  a  passenger  loeouiotive 
and  tender. 

..Ills.   Max.  stresses  in  upper  chord  =  —  57.(5,    —  103.0, 

-  13G.4,  -  155.2,  -  161.9  tons. 

Proh.  125.  A  through  Warren  truss  has  10  panels,  each 
12  feet  long  and  12  feet  deep:  find  the  niaxiiiinni  stresses 
in  tlie  ui)])er  chord  due  to  a  decapod  engine  and  tender. 

Ann.  Max.  stresses  =  - 80.8,   -145.0,   -  I'JO.U,   -217.3, 

-  225.2  tons. 

Art.  48.  Position  of  Wlieel  Loads  for  Maximum 
Moment  at  Joiat  in  Unloaded  Chord.  — I'-y  the  nUe 

deduced  in  Art.  47,  the  niaxiiuum  chord  stresses  may  be 
determined  in  the  nnlnaded  chord  of  any  siuijjle  truss,  and 
also  in  the  loaded  chord  of  s\ich  trusses  as  the  Pratt  and 
Howe,  where  the  web  members  are  vrtiad  and  inclined  so 
that  the  panel  points  of  the  upper  chord  are  directly  over 
those  of  the  lower  chord.  For  trusses  like  the  Warren  and 
lattice,  Avhere  all  the  web  members  are  inclined,  it  applies 
only  to  the  unloaded  chord.  For  the  haded  choid  of  suci 
trusses  a  modification  of  the  formula  or  rule  is  necessary, 
which  may  be  deduced  as  follows : 

Let  it  be  required  to  fiiul  the  maximum  moment  at  the 
panel  point  c  of  the  unloaded  chord,  Fig.  59. 


ii 


ttlllDGE  TItUSSES. 


167 


i  =  -83  tons 


ir)-7=-10;{.8, 

lias  10  jiaiit'ls, 

1li(>   niiixiiiniiii 

iger  locomotive 

57.G,    -103.0. 

10  i);uiels,  each 
xiiinnii  stresses 
uiil  tender. 

I'JO.y,  -217.3, 
or  Maximum 

—  I'.y  the  vn\v 
tre.sses  may  be 
in])k'.  truss,  and 
i  tlie  I'ratt  and 
(iiid  inclined  so 
re  directly  over 
the  Warren  and 
ined,  it  applies 
[  choid  of  suci 
le  is  necessary, 

moment  at  the 


Lot  W^  the  total  live  loa«l  on  the  truss,  Q  -  the  load 
(.11  the  (/(.  -  l)th  panel,  and  /•  =  the  Inul  .m  Mie  left  of 
the  (n  -  l)ti>  piiiiel. 

Let  I'  =  the  distance  of  c  from  the  left  support,  7  ^  the 
distance  be,  p  =  the  panel  length,  and  I  =  the  length  of  the 

span. 

Let  a;  =  the  distance  from  the  center  of  gravity  of  W  to 
the  right  support,  .f,=  the  distance  from  the  center  of  gravity 
.jf  F  to  the  panel  point  n,  and  x\  =  the  distance  from  Q  to 
the  panel  j/oint  n  —  1. 

The  part  of  Q  ihat  is  carried  by  the  nth  panel  point 

18  -£-^. 

Hence  the  moment  at  c  is 

I  V 

Equating  the  first  derivative  of  M  to  zero,  we  have,  since 
dx  =  dxx  =  dx.^, 


J  1>  V         ^ 


(1) 


If  the  braces  are  cipially  inclined,  that  is,  if  the  center 
(,f  moments  c  is  directly  over  or  under  the  center  of  the 
opposite  member,  as  is  usually  the  case,  we  have  |  =  ^,  and 
(1;  becomes 

P'+iQ  =  pF. (2) 

For  the  Pratt  and  Howe  trusses  7  =  0,  and  (1)  becomes 
which  is  the  same  as  the  formula  that  was  found  in  Art.  47. 


BMMll 


1G8 


nooFs  AM)  imiDaFS. 


Ft  is  convenient  in  practice  to  put  one  of  tlie  loads  at  the 
n  —  1th  panel  point,  so  that,  In  (jviwral,  wo  imist  have 


/^-f-i-Q  =  or  <^jW. 


(3) 


Proh.  126.  A  thron<,'h  Warron  truss  has  10  panels,  each 
12  fort  luui-j  and  11'  feet  deep:  tind  the  niaxiinuni  stresses 
in  tl'.c  lower  chord  due  tu  a  decajiod  eu^'ine  and  tender. 

To  tind  the  niaxitniini  stress  in  the  first  panel  2-4.  Here 
V:=y,l^\i),  and  )r=112.  Therefore  by  fonnula  (8), 
F  +  I  Q  must  1x1  erpial  to  or  less  than  o',,  X  1 12  or  5.0  tons. 
Hence  the  Ist  driver  must  be  put  at  4,  since 

I  X  8  <  5.G,  and  ^  +  \x  12.8  >  5.6. 

The  left  reaction  then  :=  8G.14  tons. 
The  moment  at  the  point  1  is 

i»/=  8G.14  X  6  -  8  X  ^»^  X  G  =  484.84. 

484.S4 


stress  in  2—1  ■ 


12 


:  40.4  tons. 


Similarly  the  maximum  stresses  in  the  other  lower  chord 
.members  are  found. 

Ma.K.  stresses  =  484.8,  111.9,  1G7.1,  203.4,  220.7  tons. 

iS'h<7.  To  lind  stress  in  4-(i,  put  i!d  driver  at  (! ;  in  0-8,  put  T.d 
driver  at  8;  in  8-U),  put  4lh  driver  at  10;  in  10-12,  put  0th  driver 
at  12. 

Art.  49.  Tabulation  of  Moments  of  Wheel  LoadB. 

—  A  diafj;niin  such  as  is  shown  in  l''i|,'  '51,  diminishes  con- 
siderably the  work  of  coinp\iting  stresses  due  to  actual  wheel 
loads.  The  Gist  diagram  is  for  an  88-ton  passenger  loco- 
motive and  its  tender:  the  se(^ond  is  fur  a  112-ton  decajtod 
engine  and  its  tender.     Aay  locomotive  can  l/e  <liagramed 


»m»to-j.-W-'i».- 


■oi^ 


le  loads  at  the 
ist  have 

...     (3) 

)  panels,  each 
iiiiuni  stresses 
i\  tender, 
el  2-4.  Here 
fornmla  (^i), 
1 2  or  5.0  tons. 


14. 


V  lower  chord 


J0.7  tons. 

in  C-Hj  put  ;;d 
,  put  0th  driver 


'heel  Loads. 

niinishi's  con- 
)  actual  wlieel 
issenger  loco- 
1^-ton  decajtod 
be  <liagramed 


nillUGE  TUUfiSES. 


1G9 


in  the  same  ninnner;  and  the  same  method  applies  to  two 
loi'oiuotivi's  coupled  together. 


.§ 

<<> 

o 

i 

o 

1 
o 

to 

o 

1 

0 

1 

o 

1 

tt> 

o 

1^     iiionperlinlt. 

a&oi     8'0  ■ -^ 

III      ^'>: 


I 


»'.§    ^!^ 


.  .  its  §1^ 

i< 5b7oni,?2:s >!  7:    ==4: 

U 64Tons;3P:o >!  1 

k 72Ton5i  37.'0  >!  1 

k eOTon5,4?:5  >,         ,     , 

K 88Ton5,47.'S '■ >i     j 

j<  - eSTonSiSa'S "•>! 


Si^ 


^  SI 

I 


I 


(a) 

I    I    »i   iS    ^ 


IS 

^      «»        Co 

cvj      tvi       SI; 


I      I 


IS      t9 


o    OQQOQ    00  6  6[ES 


I.-.- 
K— 


!  1 


U ^/?»!«  '♦.ie  ogie    i      i  !       !  !    1 

«33.67&«v?*'^*  ^4:  S^'^g  #       5l|  --is    •!§  I'l 

K     59  ?7o/75,  ?0'75 ~A       I  --^  iSlf    Si!i 

H ISTonS;  ?5:0 A  i        !  ! 

i< SZTons,  32'.5 J        I         ! 

gBTons,37:i7 ; >i  ! 

l02Ton5:42J7  ■■ •••• >«        ,     . 

/IZTonsj  47.4} >l    J 

//em5,-4St'69 - ••'•■•>) 

(b.) 
Fig.ei. 
(20  feet  =  1  inch  scale. ) 

Each  diagram  shows  the  weights  and  distances  aj)art  of 
the  wheels.  Kelow  each  Avheel,  on  the  horiznntnl  line,  is 
sliown  the  weight  of  that  wheel  togetlier  wijii  that  <*f  all 
the  preceding  ones,  and  its  distance  from  the  front  wheel. 


., 


Ma 


170 


nnoFH  Ay  I)  II  HI  DOES. 


Below  ouch  wheel  on  the  vertu  al  line  is  given  the  moment 
of  all  the  preeoding  wheels,  witli  reference  to  tluit  wheel. 
Thus,  for  the  third  driver  of  the  decapod  engine,  we  have 
4().4  tons  for  the  weight  of  it  and  the  three  preceding 
wheels,  10.5  feet  for  its  distance  from  the  front  wheel,  and 
liy5.2  foot-tons  for  the  moment  of  the  preceding  wheels  with 
reference  to  the  third  driver.  At  the  beginning  of  the 
imiforni  load  we  hav(>  112  tons  for  the  weight  of  all  the 
preceding  loads,  49.08  feet  for  the  distance  of  tliis  point 
from  the  front  wheel,  and  2910  foot-tons  for  the  moment  of 
all  the  [)receding  loads  with  reference  to  this  point.  Each 
diagram  shows  also  that,  the  itwment  at  any  ivlieel  is  equal  to 
the  moment  at  the  next  preceding  wheel  on  the  left,  plus  the 
mm  of  all  the  preceding  ivheel  loads  multiplied  by  the  dixtance 
from  the  next  left  preceding  irheel  to  the  wheel  in  question. 
Thus,  in  (a)  the  moment  of  the  first  three  wheels  about  the 
third  is  188,  and  about  the  fourth  it  is  476.  But  470  is  equal 
to  188  plus  30  multiplied  by  8;  and  similarly  for  the 
moment  with  reference  to  any  other  point.  (See  Du  Bois's 
Strains  in  Framed  Structures.) 

This  diagram  may  be  used  for  finding  reactions,  shears, 
and  monuntx.     Thus 

Let  I  =  the  length  of  the  truss,  and  the  other  notation  as 
in  Art.  4G.     Then,  from  (1)  of  Art.  4G,  the  left  reaction  is 

R  =  hPd  +  I'x  +  ^tva?), 

which  for  the  passenger  locomotive  is 
/f=.:^('2388-|-88«-h 

and  for  the  decapod  engine  is 
1 


i^"^) 


(1) 


ij- 


I 


'^no  +  U2x  +  hc3i^l 


(«) 


'**•*    -m-is 


-    '^A^  -■  ^h/c  <#^iSi, ; 


"•-■^"'"TfS,/*! 


n  the  moment 
to  that  wheel, 
giue,  we  have 
ree  preceding 
mt  wheel,  and 
ig  wheels  with 
inning  of  the 
j;ht  of  all   the 

of  this  point 
;he  moment  of 

point.  Each 
heel  is  equal  to 
e  left,  plus  the 
by  the  dititunce 
el  in  (juestiun. 
leels  about  the 
ut  470  is  equal 
ilarly  for  the 
(See  Du  Bois's 

actions,  shears, 

ler  notation  as 
ft  reaction  is 


(1) 


(2) 


niilDGK  TliUSSKd. 


171 


Suppose,  for  example,  that  the  third  tender  wheel  of  the 
decapod  engine  is  2  feet  to  the  left  of  the  right  ubutiiiciit. 
We  take  out  the  numbers  2181.8  and  102  from  the  diagram, 
and  the  reaction  is 

7i  =  ~  (2181.8 +  102 +  2). 

In  practice  it  is  convenient  to  draw  a  skeleton  outline  of 
the  truss  to  the  same  scale  as  tiie  diagram,  to  be  placed 
directly  above  it  in  the  proper  position  for  the  maximnm 
stress  in  each  member. 

Prob.  127.  If  the  span  \h  120  feet,  find  the  reactions  for 
the  passenger  Uxvinotive,  (1)  when  liie  last  tender  wlieel  l« 
."i  feet  to  the  left  of  the  right  abtttment,  (2)  when  the  UlHt 
(fewer  is  50  feet  to  the  right  ..f  the  left  abulniolil,  and 
(.^)  when  the  st>eond  pilot  wheel  is  40  feet  to  the  lidt  of  the 
right  abutiiwMit. 

Ahs.    (1)31.4  tons;   (2)  52.1  tons;   (3)  1G.4  tons. 

I\ob.  19&.  If  the  span  be  100  feet,  find  the  moments  for 
the  decapod  engine,  (1)  at  an  apex  .'SO  feet  to  the  left  of  the 
right  abutment  when  the  second  driver  is  at  that  apex,  and 
(2)  at  the  (tenter  of  the  span  when  the  first  driver  is  there. 

Ans.  (1)  1.341.4  fooUons;  (2)  1883  foot-tons. 

Prob.  129.  A  through  Pratt  truss,  Fig.  00,  has  8  jxiMels, 
each  18  feet  long  and  24  f  et  deep:  find  the  niaxinium  chord 
stresses  due  to  a  passenger  hxomotive  ami  tender  followed 
l)y  a  uniform  train  load  of  .30(M)  lbs.  per  linear  foot. 

To  find  the  position  for  the  maximum  moment  in  2  4 
caused  by  the  live  load,  we  try  the  first  driver  at  the 
[lanel  point  4.  Then  for  tbe  uniform  train  kjad  we  have 
.c  =  7  X  18  -  36  =  DO  feet.  In  this  i)Osition,  the  total  live 
load  on  the  truss 

=  88  +  1.5  X  90  =  223  tons. 


/ 


172 


HOOFS   AND   UniUGKS. 


,  ( 


Since  8  +  8  <  J  x  22.S,  aii.l  S  +  8  +  20  >  J  x  223,  this  is 
the  ciincct  position  for  niaxintiun  nionicnt  in  2-4. 
For  tl'is  loud  tlie  reaction,  hy  formula  (1),  is 

n=^\i  (2388  +  88  X  'JO  +  .75  X  Wi')  =  1 1 3.8  tons ; 

and  the  moment  at  3  is 

.V=  113.8  X  18-  188  =  1800.4. 
1800.4 


•.  max.  stress  in  2-4 


24 


=  111)  tons  =  stress  in  4-6. 


Similarly,  the  maximum  stresses  in  the  other  three  chord 
nieml)ers  arc  found  to  be  the  following : 

Max.  stress  in  3-5  =  -  128.3,  in  5-7  =  -  157.0, 
in  7-9  =  -  165.8  tons. 

Note. — To  determine  the  maxiiiiurn  strisis  in  7-0  put  10  ffPt  of  {lie 
iriiiii  to  the  lift  of  the  point  10  ;  tliiit  i.«,  put  tlic  third  tender  wheel  at 
H.     Then 

ir=  88 +  1.5x82 -211  tons. 

Rut  88  +  10  X  1.5  <  ^  X  211 ;  therefore  this  is  about  the 
correct  loading. 

!(=■•  s\i  Ci'MH  f  88  X  82  +  .75  x  82')  =  101 .7  tons. 
M  =  101.7  x  7y      (2388  f  88  x  10  +  .75  x  lo') 
=  3U70.4  foot-toiis. 


max.  stress  In  7-9  =  — 


3979.4 


=  -  lfi5,8  tons. 


Proli.  130.  A  tlirough  Tratt  truss,  Fig.  (">(»,  has  8  panels, 
each  20  feet  long  and  24  feet  deep;  find  (1)  the  maximum 
chord  stresses,  and  (2)  the  maximum  wob  stresses,  in  pach 
jKiMcl,  diw  to  a  decapod  eni^ine  and  tender  followed  bj  u 
uniform  train  load  of  3000  lbs.  per  linear  foot. 


MM 


lUUhr.K  TliUStiES. 


na 


23,  tliis  is 


tons ; 


88  in  4-G. 
luee  chord 

7.0, 


1)  fppt  of  tlie 
der  wheel  at 


about  the 


u»8. 


ns. 

5  8  panels, 

nuixituuiil 

B8,  ill  ench 

twed  \\y  a 


i 


(1)  To  find  the  position  for  niaxinunn  stress  in  2-4  wo 
try  tho  IM  driver  at  4  ;  then  x  -^  W2X>  feet;  and 

W=  11-*  +  l.r)  X  102.0  =  '2m  tons. 

Since  i  X  L'Cr,  =  3.';',  >  8  +  IL'.S,  and  <  8  +  12.8  +  12.8, 
this  is  the  eorrect  position  for  inaxinuun  .stress  ni  2-1, 
althonj^'h  if  the  third  driver  should  be  put  at  4  we  woidd 
Ket  alH)nt  the  same  stress,  as  8  +  12.8  -h  12.8  would  »v  just 
less  than  j^  of  the  load  on  the  truss.*     The  resR-tion  is 

U  =.  rJ,5  (20O'.».'.)  -I-  112  X  102.G  +  .75  X  i02.(>')  =  i:W.;5  tons. 

r.i9.3  x20-ir.2.4 


•.  max.  stress  in  2-4  = 


24 
=  stress  in  4-G. 


l-^  =  10').7  tons 


Similarly,  the  maxitnuin  stresses  in  the  other  three  chord 
members  are  found  to  be  -  181.5,  -218.9,  and  -  22(;.  I  tons. 

(•J)  To  find  the  position  for  maxinmm  shear  in  2-i}  put 
the  3d  driver  at  4;  then  *  =  100.8  feet,  and 

W=  112  4  1.5  X  100.8  =  272.2  tons. 

Since  H  f  12.8 -t- 12.1  <  It  X  272.2,  this  is  the  correct 
position  for  maximum  shear  in  2-^,     'Iho  reaction  is 

R  =-  J-  (2909.9  -f  112  X  106.8  +  .75  x  100.8')  -  140.4  tons ; 
100^ 

and  shear  in 

2-4  =  140.4  -  ^2f  (4.25  +  8.5)  - 1  x  10.5  =  131.04  tons. 

.-.  max.  stress  in  2-3  -  -  131.64  x  1.302  =  -  171.4  tons. 

•  Tho  conditions  of  Arts.  4f  and  48  may  sometimes  be  satisfied  by 
diHuruiil  iH'sllions  of  the  load. 


171 


noOFS  AM)  IlI'lOaES. 


II  I  lip  I'd  driver  lie  put  at  1  we  sliall  obtain  about  tlio 
.same  result. 

Tlio  following  stresses  are  found  in  u  manner  similar  to 
the  above : 


Muxliiiiiiii  Strcsupn  In  the  f^iitgonalg. 
2-3  =-  171.4  tons. 
3-0  =  +  130.7  tons. 
6-8  =+  !»4.7  Inns. 
7-10  =  +  0:!.(1  tens. 
0-8'  =  +  3H.3  tons. 
TAV-+    18.0  tons. 


MaxiiiMiin  Sin<«ii<>it  In  the  Vrrliiuilt. 
3-4    =+51.1  tons. 
6-0    =-  72.7  tons. 
7-8    =  -  48.9  tons. 
0-10  =  -  29.4  tons. 


Prob.  131.  A  through  ''ratt  truss  has  7  panels,  each  L'O 
feet  long  and  20  feet  deep;  the  dead  load  is  Jr>(»(>  lbs.  jier 
linear  foot,  the  live  load  is  a  i)as.seng('r  locomotive  and 
ti-mh'r  followed  by  a  uniform  train  load  of  .'iOOO  lbs.  jier 
Jinear  foot :  find  the  maximum  stresses  in  all  the  mendiers. 


CiioKo  Stressks. 

MKMlltRS. 

•.!-l,  -l-fi 

8-5 

5-T 

T-7' 

Dead  load  stnsses    .... 
Live  loatl  stri'sscs     .... 

+  48.0 
+  08.3 

-  80.0 
-167.3 

-  96.0 
-186.4 

-  96.0 
-185.4 

Ma.xiiiuiin  stresses    .... 

+  140.3 

-237.3 

-i;hi.4 

-281.4 

Stresses  in  the  Diaoonals. 


MEUIIF.U-ki 

'2-3 

8-6 

+22.6 
+64.6 

"->' 

T-6' 

Dead  Itjad  stresses  . 
Live  load  stresses  . 

-  67.9 
-138.9 

+  45.2 
+  08.7 

0.0 

+  30.5 

0.0 
+  16.7 

Maxiiniun  stresses  . 

-206.8 

+  U.I.  9 

+  87.2 

+.%.. 

-  5.9 

Ate 


nitlDGE    riWSSES. 


ITf) 


1  abuut  tho 

r  Hiiniliir  to 

the  Vtrtie.iiU. 

tons. 

tolLS. 

Ions. 

tona. 

»ls,  eacli  -'() 
[JOO  Ills.  i>or 
molivn  ami 
00  U.S.  per 
e  lucnibers. 


7-7' 

0 
4 

-  96.0 
-185.4 

4 

-281.4 

7-6' 

).0 

5.5 

0.0 
+  10.7 

i.w 

1   -  5.0 

The  livr  load  stn>ss  in  T-^V  is  +  10.7,  wliilo  tho  (load 
load  stress  in  a  s'  or  'y-it>  i.  -}•  !,"_'.»>.  Siiic»(  tlic  nifiubor 
7'-(>'  cannot  takv  (■ouij.iv.ssion,  this  losult  shows  that  tho 
("onnter  7'-(5'  is  not  uoeded  f<<r  this  loading,  tlic  meinhor 
iT-S'  taking  tho  shear  as  tension. 

STKK!<r«K>t    IN    Till'.    Vl.llTICAJ  -. 


MKMtirHK. 

8-4 

,'i-C, 

-  0.0 
-25.8 

Deatl  load  stresflcs 

Live  load  strt'Sfit's    

+  10.0 

+;io.o 

+  55.0 

-Ifl.O 
-45.7 

-01.7 

Miiviinuin  streast'S 

-25.8 

Pinh.  132.  A  through  Pratt  truss,  V\^.  .'!.'>,  has  lO  panels, 
each  L'O  feet  loiij,'  and  20  feet  deep;  the  dead  load  is  L'OOO 
Ills,  per  foot,  the  livt'  load  is  a  decapod  engine  and  tender 
followed  l)y  a  luiiforin  Irain  load  of  .'iOOO  lbs.  per  loot:  tind 
the  maxinuim  stress*  s  iu  all  the  members. 


Chord  Strewi 

'^, 

Mkmhf-rs. 

2-4,   l-Ci 

0-8 

^-10 

10-12 

u-11 

-250.0 
-|0(i.O 

Dead  load  stresses  . 
Live  lord  stresses    . 

Maximum  stresses  . 

+  00.0 

+  i«;!.o 

+ 100.0 
+  280,*! 

+  210,0 
+  ;557.7 

+  240.0 

4  :508.0 

'-253.0 

+  440.0 

+  507.7 

+  0;!8.0 

-650.0 

Stressks 

N    THE 

DlAOON.Vl-S. 

Mf.hiifbs. 

2-3 

8-C 

5-9 

7-10 

U-12 

n-10' 

«■->' 

Dcnd  load  etrestos  .... 
Live  loiMl  stresses  .... 

-127.3 
-2.30.6 

+  98.9 
+  18fl..'i 

+  70.7 
+146.9 

+  42.4 
+111.4 

+  14.1 
+S0.2 

0.0 
+54.1 

(1.(1 
+33.1 

Mn.xlinmn  stresops ... 

-357.S 

42W.4 

+217.6 

+i.-ins 

+<>'  3 

+40.0 

-9.3 

\ 


mvi* 


IMAGE  EVALUATION 
TEST  TARGET  (MT-3) 


k{0 


// 


^^, 


Ik?   MJi 


.<'    M? 


<[<■  -k' 


%"*" 


i<'^ 


Ui 


f/. 


1.0 


I.I 


1.25 


PIIIIIM    12.5 
Z  1112 


IM 

m 


1.4 


20 

i.8 


1.6 


i 


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O^ 


-^,^..-.  ...-.•   1  -,^.-  ..i;,: 


17<! 


ROOFS  AND  IIRWGES. 


Streshes  in  the  Verticals. 


Mkmiiriui. 

8-1 

6-6 

--3 

£-1(1 

11-12 

Dead  load  stresses  .  .  . 
Live  load  stresses    .  .  . 

+20.0 
+  51.1 

-  50.0 
-103.9 

-  30.0 

-  78.8 

-10.0 
-50. 7 

-  0,0 
-38.3 

Maximum  stresses  .  .  . 

+  71.1 

-153.0 

- 108.8 

-60.7 

-38.3 

Since  the  live  load  stres.s  in  11-10'  or  11-10  is  -f  54.1, 
while  the  dead  load  stress  in  the  sa'ne  diagonals  is  —  14.t, 
the  maximum  stress  is  the  ditYerence,  or  -f-40.0.  Alsc, 
since  the  live  load  stress  in  9-8'  or  h-H  is  -f  3.3.1,  while  the 
dead  load  stress  is  —42.4,  the  maxinuni  stress  in  5)'-8'  or 
9-8  is  —  i).3.  Hence  counters  are  needed  only  in  the  two 
middle  panels.  ,  • 

Prob.  133.  A  through  Warren  truss  has  8  panels,  each  W 
feet  long  and  15  feet  deep;  the  dead  load  is  2000  IIks.  jyer 
linear  foot,  the  live  lo.r,i  is  a  passenger  locomotive  and 
tender  followed  by  a  uniform  train  load  of  ,'5000  lbs.  per 
linear  foot:  find  the  maximum  stresses  in  all  the  member.s. 

Prob.  134.  A  deck  Pratt  truss  has  10  panels,  each  20  feet 
long  and  24  feet  deep ;  the  dead  load  ij  given  by  forimda  (1), 
Art.  15,  the  live  load  is  a  decapod  engine  and  tender  followed 
by  a  uniform  train  load  of  3000  lbs.  per  linear  foot :  -find 
the  maximum  stresses  in  all  the  members. 


I  Ko*  it^tHiJMmmm 


■Mi  i,iwiiwi>m»iiiwiiiiwffw»imiiii<j<«i 


C-IO 

11-12 

-10.0 

-50.7 

-  0,0 

-38..S 

-60.7 

-38.3 

.0  is  +  54.1, 
Is  is  -  14.t, 
40.0.  Alsc, 
.1,  while  the 
5S  in  5)'-8'  or 
T  in  the  two 

nels,  each  lo 
iOOO  lh.s.  i)er 
)inc»tive  aixl 
000  lbs.  per 
he  members. 

each  20  feet 
formiila  (1), 
tier  followed 
it  foot :  "find 


7l 


■ij-     *-  If 


Ik  I 


It 


CHAPTER  IV. 

MISCELLANEOUS  TRUSSES. 
Roof  Trusses. 

Art.  SO.  The  King  and  Queen  Truss  — The  Fink 

«rru9g._.The  types  of  roof  trusses  most  commonly  used  for 
spans  from  30  feet  to  100  feet,  and  even  to  130  feet,  are 
shown  in  Chapter  I.  The  King  and  Queen  truss.  Fig.  7,  is 
a  common  form  for  wooden  trusses,  or  for  trusses  that  are  all 
of  wood  except  the  verticals,  which  are  iron  or  steel  tie-rods. 
This  type  of  truss  is  sometimes  used  also  when  it  is  entirely 
made  of  steel. 

The  Be^van,  or  Fink  roof-truss.  Fig.  10,  is  a  very  common 
and  economical  type  of  truss  for  spans  up  to  130  feet.  The 
struts,  3^,  5-6,  7-8,  are  normal  to  the  rafter,  dividing  it 
into  equal  parts,  and,  with  the  upper  chord,  are  made  either 
of  wood  or  iron.  The  other  members  are  ties,  and  are  made 
of  iron  or  steel.  This  type  of  truss  is  often  entirely  of  steel, 
and  is  commonly  used  for  iron  or  steel  roofs  over  mills, 
shops,  warehouses,  train-sheds,  etc.  Some  of  the  largest 
triiss.'s  of  this  type  are  those  in  the  car-slops  of  the  Tenn- 
sylvania  railroad  at  Altoona,  ?a.,  having  a  span  of  132  feet. 

Tilt'  slope  of  the  rafter  is  usually  determined  by  the  kind 
of  roof  covering  used.  Slate  should  not  be  used  on  a  roof 
when  the  slope  is  less  than  1  vertical  to  3  horizontal,  and 
preferably  1  vertical  to  2  horizontal.  Gravel  should  not  lie 
used  on  a  slope  greater  than  1  vertical  to  4  horizontal.     Tin 

177 


ii 


178  HOOFS  AND  nUWGES. 

may  bo  used  on  any  slope,  or  on  a  flat  roof.  Corrugated 
iron  should  not  be  used  on  a  slope  less  than  1  to  3;  for 
flatter  roofs  than  1  to  3,  of  corrugated  iron,  are  liable  to 
leak  under  a  driving  rain  as  the  usual  joints  are  not  tight. 
When  possible  the  slope  should  not  be  less  than  1  to  2. 

The  following  table  gives  the  approximate  weight  per 
square  foot  of  roof  coverings,  exclusive  of  steel  construction. 

Approximate  Weight  per  Square  Foot  oit  Uoof  Covbrikos. 

Corrugatedlron,  unbearded,  No.  26  to  No.  18.     .     .     •  1  to  3  lbs. 

Felt  and  asphalt,  without  sheathing ^  "^s. 

Felt  and  gravel,  without  sheathing 8  to  10  lbs. 

Slate,  without  sheathing,  -A"  to  J" '  **  ®  '*'^- 

Copper,  without  sheathing 1  to  1 J  lbs. 

Tin,  without  sheathing 1  to  1}  Ib.s. 

Shingles,  with  lath ^i  '•'s- 

Skylight  of  glass,  ^\"  to  i",  including  frame    ....  4  to  10  lbs. 

White  pine  sheathing,  1"  thick 3  lbs. 

Yellow  pine  sheathing,  1"  thick 4  lbs. 

Spruce  shoathing,  1"  thick 2  lbs. 

Lath  and  rlaster  ceiling 8  to  10  lbs. 

'Pile  flat 16  to  20  lbs. 

Tile!  corrugated «  *»  !<>  1^«- 

Tile,  on  3"  fireproof  blocks 30  to  36  lbs. 

The  weight  of  the  steel  roof  construction  must  be  added 
to  the  above.  For  ordinary  light  roofs  without  ceilings, 
the  weight  of  the  steel  construction  may  be  taken  at  5-lbs. 
per  square  foot  for  spans  up  to  60  feet,  and  1  lb.  additional 
for  each  10  feet  increase  of  span.  (Manual  of  useful  infor- 
mation and  tables  appertaining  to  the  use  of  Structural 
Steel,  as  manufactured  by  the  Tassaic  Rolling  Mill  Co., 
Paterson,  N.J.     By  George  H.  Blakeley,  C.E.) 

Prob.  135.  A  truss.  Fig.  02,  with  one  end  free,  has  its 
span  120  feet,  its  rise  30  feet,  the  ties,  3-4,  5-6,  7-8,  9-10, 


M  ISC  EL  I A  N  i:0  US   TR  USSE8. 


179 


Corrugated 
1  to  3;  for 
ro  liable  to 
e  not  tight. 
lto2. 
weight  per 
onstructiou. 

CovuniNos. 

1  to  3  lbs. 

2  lbs. 
8  to  10  lbs. 

7  to  9  lbs. 

1  tolill^s- 

1  to  11  lb.s. 

2^  lbs. 

4  to  10  lbs. 

3  lbs. 

4  lbs. 
2  1b,s. 

8  to  10  lbs. 
16  to  20  lbs. 

8  to  10  lbs. 
30  to  35  lbs. 

ist  be  added 
out  ceilings, 
len  at  5- lbs. 
b.  additional 
useful  infor- 
f  Structural 
ng  Mill  Co., 

free,  has  its 
.6,  7-8,  9-10, 


11-12,  vprtical,  dividing  the  rafter  into  (5  ecjual  parts;  the 
dead  load  i)or  panel  is  2.5  tons,  snow  load  per  panel  l.fi  tons, 
normal  wind  load  per  i)anel,  wind  on  fixed  side,  2  tons :  find 


all  the  stresses  in  all  the  members  of  the  half  of  the  truss 
on  the  windward  side. 

^  Stresses  is  the  Loweu  Chord. 


Mf.mukrh. 

2  6 

c-s 

8- in 

10-12 

12-14 

Dead  load  stresses  .  .  . 
Snow  load  strcssi's  .  .  . 
Wind  load  stresses  .  .  . 

+  27.5 
-16.5 
+  17.8 

+  25.0 
+  15.0 
+ 15.6 

+  22.5 
+  1.3.5 
+  13.4 

+20.0 
+  12.0 
+  11.1 

+  17.5 
+  10.5 
+  8.9 

Maximum  stresses  .  .  . 

+61.8 

+56.6 

+  49.4 

+43.1 

+36.9 

Stresses  in  the  Upp  ;r  Chord. 


Mrmiiera. 

2-3 

3-6 

6-7 

7-0 

8-11 

11-18 

Dead  load  stresses 
Snow  load  stresses 
Wind  load  stresses 

-.30.75 
-18.45 
-14.62 

-27.05 
-16.77 
-13.00 

-21.15 
-16.09 
-11.48 

-22.35 
-l.'Ul 
-  9.96 

-10.55 
-11.73 

-  8.44 

-16.75 
-10.05 
-  7.44 

Maximum  stresses 

-63.72 

-5772 

-61.72 

-45.72 

-.30.72 

-.34.24 

ii'^' 


»*ffir  JMItilfcgya^g'rT  ■ 


I 


180 


ROOFS  AND  n  HI  DOES. 


fTBKHSES    IN    TIIK    VeHTICALS. 


Memhkks, 


Dead  load  Hlresflo.s  .  . 
Snow  load  slressfS  .  . 
Wind  load  stresses    . 


Maximum  streases 


f>-« 


+  1.25 
+0.76 
+  1.12 


+  3.12 


7-8 


+  2.50 
+  1.60 
+  2.24 


+0.24 


9-10 


+  3.76 
+2.26 
+  3.30 


+  9.36 


11-12 


+  5.00 
+  3.00 
+  4.46 


+  12.46 


13-14 


+ 12.60 
+  7.60 
+  6.56 


+25.56 


(3-4  is  not  necessary  to  tlie  stability  of  the  truss.) 
Stresses  in  the  Diagonals. 


MRMItRKK. 

8-6 

6-8 

7-10 

0-12             11-14 

Dead  load  stre-sses  . 
Snow  load  stresses  . 
Wind  load  stresses  . 

-2.80 
-1.68 
-2.48 

-3.5r 

-2.11 
-3.16 

-  4.50 

-  2.70 

-  4.04 

-  5.66 

-  3.33 

-  5.00 

-  6.70 

-  4.02 

-  0.00 

Maximum  stresses  . 

-0.96 

-8.79 

-11.24 

-13.88 

-16.72 

Prob.  136.  A  truss  of  the  type  oi  fig.  62,  with  one  end 
free,  has  its  span  150  feet,  its  rise  25  feet,  the  ties,  3-4, 
5-6,  etc.,  vertical,  dividing  the  rafter  into  six  equal  parts; 
the  dead  load  per  panel  is  2.8  tons,  snow  load  per  panal 
1.5  tons,  normal  wind  load  per  panel,  wind  on  fixed  side, 
1.8  tons :  find  all  the  stresses  in  all  the  members  of  the  half 
of  the  truss  on  the  windward  side. 

Stresses  in  the  Lower  Chord. 


Mkmhkks. 

2-n 

fl-8 

8-10 

1(V12 

12  14 

Dead  load  stresses  .  . 
Snow  load  slres.ses  .  . 
Wind  load  stresses  .  . 

+40.20 
+  24.76 
+  22.79 

+42.00 
+22.50 
+  19.93 

+37.80 
+  20.26 
+  17.00 

+  3.S.00 
+  18.00 
+  14.20 

+29.40 
+  16.76 
+  11.34 

Maximum  8t.re^^ses   .  .  j  +93.74 

+  84.43 

+  75.11 

+06.80 

+  66.49 

mmm 


i 

13-U 

00 
00 
46 

+ 12.60 
+  7.50 
+  6.56 

46 

+25.56 

•i 

11-14 

66 
33 
.00 

-  6.70 

-  4.02 

-  6.00 

.88 

-16.72 

ith  one  end 
lie  ties,  3-4, 
Bqual  parts; 
d  per  par.al 
1  fixed  side, 
s  of  the  half 


V12 

12  14 

53.60 
8.00 
4.20 

+29.40 
+  16.76 
+  11.34 

16.80 

+  56.49 

WittM 


MISCKLLA  NKOUS    Tit  USSES. 
Strbssbs  in  the  UrPEU  Choki). 


181 


Mkmiikuh. 

Duiwl  load  slresseH  . 
Snow  load  stresses . 
Wind  load  st'esses. 

2-i 

!Wi 

B-T 

T-it 

0-11 

11-18 

-48.97 
-26.24 
-20.74 

-44.62 
-23.86 
-18.32 

-80.69 

-40.07 
-21.47 
-15.91 

-86.62 
-19.08 
- 13.50 

-31.16 
-16.70 
-11.09 

-26.71 
-14.31 
-  8.96 

Maximum  stresses  . 

-95.96 

-77.46 

-68.20 

58.96 

-49.98 

Stresses  in  the  Verticals. 


Mkuiikks. 

B-fi 

t-'» 

r-io 

11-12 

18-14 

Dead  load  stresses  .  .  . 
Snow  load  stresses .     . 
Wind  load  stresses .  .  . 

+  1.40 
+  0.76 
-J-0.04 

+2.80 
+  1.50 
+  1.89 

+  4.20 

+2.25 
+2.84 

+  6.60 
+  3.00 
+  3.78 

+  17.02 
+  9.12 
+  6.76 

Maximum  stresses  .  .  . 

+3.09 

+6.19 

+9.29 

+ 12..38 

+31.90 

Stresses  ^s  the  Diaoonai-s. 


Memherh. 

8-0 

.'.-S 

7-10 

9-12 

11-14 

Dead  load  stresses  .  .  . 
Snow  load  stresses.  .  . 
Wind  load  stresses.  .  . 

-  5.16 

-  2.76 

-  3.38 

-  5.60 

-  3.00 

-  3.82 

-  0.27 

-  3.30 

-  4.27 

-  7.28 

-  3.90 

-  4.97 

-  8.  .34 

-  4.47 

-  6.65 

Maximum  stresses  .  .  . 

-11.29 

-12.42 

-13.90 

-16.15 

-18.40 

Prob.  137.  A  Fink  truss,  Fig.  10,  with  one  end  free,  has 
its  span  150  feet,  its  rise  25  feet,  the  rafter  divided  into 
four  equal  parts  by  struts  drawn  novii'.al  to  it,  the  dead  load 
per  panel  3  tons,  the  snow  load  per  panel  1.5  tons,  the 
normal  wind  load  per  panel,  wind  on  fixed  side,  2  tons: 
find  all  the  stresses  in  all  the  members  of  the  half  of  the 
truss  on  the  windward  side. 


■; 


^^sm 


MaSitWMili^ 


J 


182 


R00F8  AMJ  BHWaES. 
Sthkshks  in  tiik  Lower  Chord. 


Mehbkiui. 

'.'-I 

4-6 

ft-m 

Dead  load  streHses    .... 
Snow  load  »trc88e»    .... 
Wind  load  stressos   .... 

+  31.50 

+  15.76 
+  16.84 

+  27.00 
+ 13.60 
+  12.68 

+  18.00 
+  9.(H) 
+  o.:J4 

Maximum  stressuH    .... 

+03.09 

+  53.18 

+;}3.34 

Stiikssks  in  the  Ui'Pkr  Ciiouu. 


Mkuderh. 

2-3 

«-a 

6-T 

7-9 

Dead  load  atresseH   .... 
Snow  load  stresses  .... 
Wind  load  stresses  .... 

-33.15 
-10.58 
-14.30 

-32.22 
-10.11 
-14.30 

-31.29 
-15.06 
-14.30 

-30.30 
-16.18 
-14.30 

Maximum  stresses  .... 

-04.09 

-62.09 

-01.30 

-69.90 

Stressks  in  tiik  VVkii  Mkmhers. 


Mkmiikks. 

8-t,  7-3 

£•-!) 

4-5,  6-S 

c-s 

e-9 

Dead  load  stresses  .  .  . 
Snow  load  stresses .  .  . 
Wind  load  stresses .  .  . 

-2.85 
-1.43 
-2.00 

-  5.70 

-  2.86 

-  4.00 

+4.60 
+  2.25 
+  3.12 

+-  9.00 
+  4.50 
+-  6.32 

+  13.50 
+  0.75 
+  0  48 

Maximum  stresses . 

-0.28 

-12.66 

+  9.87 

+  19.82 

+  2ft  73 

Art.  51.  The  Crescent  Truss,  Figs.  63  and  M,  is  a 
good  form  for  comparatively  large  spans.  Riveted  iron- 
work is  used  throughout. 

Prob.  138.  A  circular  Crescent  truss,  Fig.  0.3,  with  one 
end  free,  lias  its  span  160  feet,  the  rise  of  the  upper  chord 
32  feet,  the  rise  of  the  lower  chord  20  feet,  the  number  of 


6-10 

+  18.00 
+  0.(H) 
+  tt.U 

+;w.34 

7-9 

9 

-3(>.3(» 

6 

-10.18 

0 

-14.30 

0 

-59.90 

-s 

fi-9 

9.00 
4.50 
8.32 

+  13.50 
+  0.75 
+  0  48 

9.82 

+  29.73 

md  M,  is  a 
iveted  iron- 

53,  with  one 
upper  cliortl 
!  numher  of 


I 


MI8CKLLANK0U8  TRUSSES. 


1S3 


l)aiiels  8,  the  apexes  of  both  the  ui)per  and  lowpr  chords 
lying  on  arcs  of  circles,  dividing  thcin  into  8  eqna,\  parts ; 
the  dead  load  is  3  tons  per  panel,  the  wind  load  is  2  tons 


iTiu.oa 


per  panel,  the  wind  being  supposed  to  act  vertically  on  the 
fixed  side  only : '  find  all  the  stresses  in  all  the  members  of 
the  half  of  the  truss  on  the  windward  side. 

The  computation  of  the  stresses  is  effected  by  the  appli- 
cation of  the  principles  of  Chapter  I. 

The  radius  of  the  upper  chord  =  IIG  feet. 

The  radius  of  the  lower  chord  =  170  feet. 

The  lengths  of  2-3,  3-5,  etc.     =  22.04  feet. 

The  lengths  of  2-4,  4-6,  etc.     =  20.81  feet. 

The  horizontal  distance  of  each  apex  from  the  left  abut- 
ment 2,  and  its  vertical  distance  above  the  line  2-2',  are  as 

follows : 

(S)        (5)  (7)  (9) 

17.33,  36.94,  68.06,  80  feet. 

13.01,  23.70,  29.91,  32  feet. 

(4)        (6)        (8)         (10) 
18.93,    38.76,    69.22,    80  feet.    , 
8.66,    14.92,    18.72,    20  feet. 


Hor.  distance  from  2, 
Height  above  line  2-2', 


Hor.  distance  from  2, 
Height  above  line  2-2', 


1  n  the  normal  wind  pressure  were  taken,  it  would  have  a  different 
value  for  each  panel,  owing  to  the  curved  surface  of  the  rafters,  which 
wonid  increase  the  difficulty  of  the  computation. 


1H4 


ItooFS  AM)  nniDGES. 


Tlie  dead  load  reiu-tion  =  10.fi  tons. 

Tho  stresses  in  tlie  elunds  are  best  found  by  inonientH; 
for  the  other  nienihers  the  method  of  resolution  of  forces 
may  be  used. 

Thus,  to  compute  the  dead  load  stress  in  2-il,  we  find  the 
lever  arm  of  2-3  to  be  4.89  feet. 

.-.  dead  load  stress  in  2-^1  =  -  lM>^«i?§  =  _  40.7  tons. 

4.oy 

Similarly,  the  stress  in  2-4  =  IM  >ill:33  ^  g^  ^  ^^^^ 

To  find  the  stresses  in  3-4  and  3-5  pass  a  secition  cutting 
2-4,  ;i-4,  and  3-5.  Then,  denoting  the  stresses  in  3-4  and 
.'5-5  by  s,  and  .%  we  have,  for  horizontal  and  vertical  com- 
ponents respectively, 

35.1  X  .9095  + A-^ s,  +  .889  »,  =  0, 

yjlS'  +  4.96* 

and       35.1  x  .4150  -  ^  ».  +  .4579  s,  +  7.5  =  0. 

.•.  s,  =  +   5.0  tons  =  stress  in  3-4 ; 
and  Sj  =  —  37.7  tons  =  stress  in  3-6. 

The  wind  load  reaction  =  5,1  tons. 

6.1  X  18.93 


wind  load  stress  in  2-3  =  - 


4.89 


=  - 19.7  tons. 


In  this  way  all  the  stresses  may  be  found. 
Stresses  in  the  Top  Chord. 


MiMHKRH. 

2-8 

8-5 

5-7 

T-9 

Dead  load  stres-ses  .  .  . 
Wind  load  stresses .  .  . 

-40.7 
-19.7 

-37.7 
-17.0 

-37.8 
-16.6 

—37.8 
-14.8 

Maximum  stres-ses  .  .  . 

-60.4 

-54.7 

-54.4 

-52.6 

■<Wi1l»JI.I|JJJ! 


iiioiuout.s ; 
of  force.s 

'e  find  the 

40.7  tons. 

3.1  tons. 

on  cutting 
n  3-4  and 
tical  com- 

I), 


19,7  tons. 


MISCELLANEOUS  TltUSSES. 


Rtuksheh  in  Tin;  Hottom  TiioRn. 


185 


Mkmhei». 

i-i 

4-« 

6-s 

»-10 

Dead  load  stresses  ,  .  . 
Wind  load  Rtrosses .  .  . 

+  35.1 
+  16.0 

+  30.9 
+  10.4 

+  37.8 
+  16.0 

+  38.1 
+  12.7 

Maximum  stresses  .  .  . 

+  51.1 

+  63.3 

+  52.8 

1  50.8 

Stresses 

IN    THE 

Weii 

Memiieus. 

MKMhBRB. 

84 

■^fl 

T-S 

U-li) 

1-5 

(17 

8-9 

Dead  load  stresses 
Wind  load  stresses 

+  5.0 
+  2.2 

-1.8 
+  2.0 

+  5.0 

+  2.8 

-1.4 

+  1.4 

+  5.1 
-2.0 

-0.6 
-2.0 

+4.2 
+2.4 

Max.  compression 
Max.  tension  .  .  . 

+0.0 

+7.2 

-1.8 
+0.2 

+  0.0 

+7.8 

-1.4 
+0.0 

+0.0 
+5.1 

-2.6 
-0.0 

+  0.0 
+0.6 

Prob.  139.   A  circular  Crescpnt  truss,  Fig.  64,  with  one 
end  free,  has  its  span  160  feet,  the  rise  of  the  upper  chord 


zriK.tM 


32  feet,  the  rise  of  the  lower  chord  20  feet,  the  apexes  of 
both  the  upper  and  lower  chords  lying  on  arcs  of  circles. 
The  upper  chord  has  8  pu.^els,  all  of  the  same  length.  The 
lower  chord  has  7  panels ;  each  of  the  5  panels  4-6,  6-8, 


ss@%«ae!#i 


18G 


ItOOFS   AM)   ItlllhdKS. 


«>,..... 


^ 


etc.,  is  o(iual  lo  one-eighth  of  the  lower  cliord,  while  the  two 
end  puiu'lH,  L'-l  luul  2-4',  arc  eacii  e(|iial  to  IJ  of  the  other 
panels,  tliat  's,  v(\\\n\  to  1  \  ei{,'lith,s  of  the  Iowit  chord  ;  liie 
(lead  load  is  .'t  tons  per  panel  inid  the  wind  load  is  12  tons 
per  panel,  the  wind  hei'ig  snpposed  to  act  vertically  on  the 
fixed  side:  lind  all  the  .stresses  ii.  all  the  menihers  of  the 
lialt  of  the  trnss  on  the  windward  side. 

Here  the  radii  of  the  npjter  and  lower  chords  are  110  and 
170  feet,  respectively,  as  in  i'rob.  138,  and  the  eoiVrdinates 
of  the  apexes  of  the  upper  chord  are  also  the  same  as  in 
I'roW.  l;i8. 

The  length  of  2-1  is  found  to  be  31.21  feet,  and  the  cottr- 
di nates  of  4  to  be  28.77  feet  and  12.0'.)  feet. 

The  reactions  are  the  same  as  in  I'roVj.  138. 


Di'ad  load  stress  in  2-3  =  — 


Dead  load  stress  in  2-4  = 


Wind  load  stress  in  2-3  =  — 


Wind  load  stress  in  2-4  = 


lO.fi  X  28.77 
8.26  '  " 

lO.f)  X  17.33 

6.83 

6.1  X  28.77 
8.26 

6.1  X  17.33 

5.83 


-  3().G  tons. 

31.2  tons. 
- 17.7  tons. 

16.1  tons. 


Strebseh  in  tub  Tor  Chord. 


Memiiiim. 

^-A 

8-8 

M 

T-» 

Dead  load  stresses 

Wind  load  stresses 

-30.0 
-17.7 

-40.5 
-18.4 

-30.3 
-10.8 

—38.7 
-14,4 

Maxinuun  stresses 

-54.3 

-58.9 

-56.1 

-53.1 

-  •^•maffrnm 


7-» 

3 
8 

-38.7 
-14,4 

1 

-53.1 

MlSVFf.  I.  .1 S  Kors   Til  I '  ^  SA'.S. 


187 


Sl'IH-'.^HKH    IN    TIIK    liorTOM    ('lll)ltll 

MlMHRIW. 

!i-l 

4-(i 

ft-s 

H-h' 

+  ::rt.o 

+  !•-'. 4 
+  10.0 

Dnul   lliilll  Htl'I'MMCH 

Wind  Iciail  utri'HMiH 

+  31.2 
+  16.1 

+  35.1 
+  10.2 

+  :wi.o 

+  14.7 

.Miixiiiiuin  KtrcHHts 

+  46.3 

+  51.3 

+  51.3 

Strksuks  in  TIIK  Wkii  Mkmiikiih. 


Mkhhkkh. 

11-1 

4-.'. 

K-ll 

<:-7 

"-i 

s-!> 

Doail  load  stresses    .  . 
Wind  loHil  streBst'H    ■   . 

+  7  2 
+  3.0 

+3.2 
+  1.4 

+4.1 
+  2.2 

+  2.7 
+  2.0 

+  3.1 

+  2.0 

+0.0 

H3.0 

+2.7 

+  5.7 

Maxinniiii  stmnes .  .  . 

+  10.2 

+4.0 

+0.3 

+  6.  ft 

BiiiDOK  Tbussks. 

Art.  52.  The  Fegram  Truss  -  The  Parabolic  Bow- 
string Truss.  —  Tli«  I'ognim  truss,  Fig.  (i."i,  (ionsists  of  U»e 
same  nuiuber  of  panels  in  each  cliord.     Ml  the  panels  of 


oiich  chord  are  of  equal  Ipngth,  the  upi)or  chord  panels 
being  shorter  than  the  lower.  The  apexe.s  of  the  upper 
chord  lie  on  an  arc  of  a  cinrle,  the  chord  of  which,  .V.'V,  is 
about  one  and  one  half  panel  lengths  shorbu'  than  the  span. 
The  rise  of  the  <ipper  chord  is  such  as  to  make  the  posts, 


ifcrtrii?  ni^iiliiiitii 


183 


ROOFS  AND  llltlDGES. 


4-6,  6-7,  8-9,  nearly  equal  in  length,  or  it  may  be  so  taken 
that  the  posts  will  decrease  in  length  toward  ♦^he  ends.  In 
a  deck  bridge  the  upper  chord  is  made  straight  and  the 
lower  chord  curved. 

Prob.  140.  A  through  Pegram  truss.  Fig.  65,  has  a  span 
of  200  feet,  divided  into  7  panels,  each  28.57  feet  long;  the 
upper  apexes  lie  on  an  arc  of  a  circle,  the  center  height 
being  15  feet  above  the  chord  3-3',  which  is  160  feet  long; 
the  upper  panels  are  all  of  equal  length,  the  members  2-3, 
4-5,  6-7,  8-9  are  struts,  all  the  remaining  web-members  are 
ties,  the  dead  and  live  loads  are  10  tons  and  18  tons  per 
panel  per  truss:  find  the  stresses  in  all  the  members. 

Here  the  radius  of  tlie  upper  chord  —  220.83  feet. 

The  lengths  of  the  upper  panels  3-5, 5-7,  etc. =23.37  feet. 

The  horizoatal  distance  of  each  upper  apex  from  the  left 
.abutment  2,  and  its  vertical  height  above  th  to,ver  chord 
2-2',  are  the  following : 

(3)       (5)        (7)       (9) 

Kor.  distance  from  2,  20.00,  42.20,  65.05,  88.05. 

Height  above  lower  chord,  24.00.   31.29,  36.23,  38.65. 

The  dead  load  reaction  =  30  tons. 
Then,  dead  load  stress  in 


4-6  = 


30  X  42.20  - 10  X  13.63 


31.29 
Also,  d^ad  load  stress  in  9-9' 

•     30  X  4  X  28.57  -  60  X  28.57 


=  36.1  tons. 


38.65 
Similarly,  dead  load  stress  in  3-5 


■  44.3  tons. 


30x28.57 
25.47 


=  -  33.6  tons. 


Hi-WIW 


(9) 
,  88.05. 
,  38.65. 


MISt'KLL .  1  -V  EO  UH  Tit  USSES. 


189 


To  find  the  dead  load  stress  in  any  web  member,  as  5-6, 
pass  a  section  cutting  5-7,  5-6,  and  4-6,  and  take  moments 
around  the  intersection  of  5-7  and  4-6,  which  is  102.53  feet 
to  the  'eft  of  2. 


.-.  dead  load  stress  in  5-6 

30x102.53-10x731.1 


=  12.2  tons. 


"  1'44.1 

The  stresses  in  the  other  web  members  are  found  in  like 
manner. 

The  live  load  stresses  in  the  chords  and  in  2-3  and  3-4 
are  a  maximum  for  a  full  load,  and  may  therefore  hi 
i  und  by  multiplying  the  corresponding  dead  load  stresses 
l)y  1.8. 

For  a  maximum  in  5-6  and  4-5  the  live  load  covers  all  the 
joints  from  the  right  to  6.     The  reaction  for  this  loading  is 

ij    .M(l  +  2-|-3  +  4  +  5)  =  ^^^ton8. 

.-.  live  load  stress  in  5-6  =  ^^4^^  X  ^^  =  27.4  tDus. 

7  144.1 

For  a  maximum  in  9-8'  the  joints  4',  6',  and  8'  are  loacled. 
The  reaction  for  this  loading  is 

/J  =  1^  (1+- 2  4- 3)  = -4^  tons, 

which  is  also  the  maximum  shear  in  this  panel.  Since  1;he 
uppfer  and  lower  c><.ord  members  of  this  paiiel  are  horizontal, 
the  stress  in  9-8'  is  found  by  Art.  18. 

.-.  max.  stress  in  9-8'  =  ^  x  ^^-  =  18.5  tons. 

7        oo.oo 


190 


ROOFS  AND   li  HI  DUES. 


Sthkssks  IV  THE  Upper  Ciiohd. 


Mkmbkkh. 

!>.') 

5-7 

7-9 

9-U' 

Dead  lofwl  stresseH 

Live  load  stresses     

-S3.6 
-00.4 

-  42.4 

-  70.2 

-  45.1 

-  81.1 

-  44.3 

-  70.7 

Maximum  stresses 

-94.0 

-118.6 

-  120.2 

-  124.0 

STnESSBS  IN  TUB    LoWEK  ClIOIlD. 


Memiiekh. 

2-4 

4-0 

C-i 

8-8' 

Dead  load  stressep 

Live,  loiul  stresses 

+25.0 
+  45.0 

+  30.1 
+  65.0 

+  41.6 
-t    74.0 

+  44.0 
+  80.3 

Maxi.iuun  stresses 

+  70.0 

+  101.1 

+  116.6 

+  124.0 

II 


Stresses  in  the  Web  Ties. 


Members. 

8-4 

e-« 

7-3 

9-8' 

»'-0' 

Deiid  load  stresses  .  . 
Live  load  stresses    .  . 

+20.6 
+37.2 

+  12.2 

+  27.4 

+  6.2 

+  22.7 

+  0.0 

+  18.6 

+  0.0 
+  14.1 

Muximum  stresses  .  . 

+67.8 

+30.0 

+28.0 

+  18.5 

+  14.1 

Stresses  in  the  Web  Struts. 


Mkmbebs. 

2-8 

4-6 

6-7 

8-9 

Dead  load  stresses 
Live  ioad  stresses  .  .  . 

-  38.8 

-  09.8 

-10.0 

-27.8 

-   1.3 
-17.3 

-  0.0 
- 12.0 

Maximum  stresses   .  . 

-108.6 

-38.4 

-18.0 

-12.0 

m^Km 


+  0.0 
+  14.1 

+  14.1 


8-9 


-12.0 


MISCELLANEOUS   THUSfitS. 


191 


See  Framed  Structures,  by  Johnson,  Bryan,  iiinl  Turneaure  j  also 
Engineering  News,  Dec.  10  and  17, 1887,  and  Feb.  11,  1801,  where  the 
answers  differ  very  slightly  from  the  above  on  account  of  the  center 
member  9-9'  being  t^  less. 

Prob.  141.  A  through  parabolic  bowstring  triisfj,  Fig.  (Jfi, 
has  8  pauels,  each  24  feet  long,  and  32  feet  center  depth, 


the  verticals  are  ties  and  the  diagonals  are  struts ;  the  dead 
and  live  loads  are  10  tons  and  15  tons  per  panel  per  truss : 
find  the  stresses  in  all  the  members. 

SxRr^SSES    IN   THK   TOP   ClfORD. 


Memiiers. 

2-3 

-  65.0 

-  97.6 

ft  T 

7  9 

Dead  load  stresses    .... 
Live  load  stresses 

-  69.4 
-104.1 

-  61.8 
-  92.7 

-  60.2 

Maximum  stresses    .... 

-173.6 

-102.5 

-154.6 

-160.5 

Dead  load  stress  in  each  panel  of  lower  chord  =  +■  60  tons. 
Live  load  stress  in  each  panel  of  lower  chord  =  -|-  90  tons. 
Maximum  stress  in  each  panel  of  lower  chord  =  -1- 150  tons. 


Stresses 

IX   THE 

Diagonals. 

Mkmiikrs. 

r-a 

5-S 

7-10 

4-8 

6-7 

0.0 
-18.1 

Dead  load  stresses . 
Live  load  stressi-s  . 

0.0 
-13.0 

0.0 
-16.9 

0.0 
-18.0 

0.0 
-1.5.9 

0.0 
-18.0 

Maximum  stresses . 

-13.0 

-15.9 

-18.0 

-16.0 

-18.0 

-18.1 

uf 


fSS^- 


I,  i 


192 


liOOFS  AND  UHIDGES. 
8tke88k»  in  tub  Vkuticals. 


! 


Mkmiikkh. 

3-1 

5-8 

T-9 

9-10 

Dead  load  strcssi'S  .... 
Live  load  stresses    .... 

+  10.0 
+  15.0 

+  10.0 
+  19.7 

+  10.0 
+22.6 

+  10.0 
+23.4 

Maximum  stresses  .... 

+26.0 

+  i!0.7 

+32.6 

+33.4 

Art.  53.  Skew  Bridges  ave  those  in  which  the  end 
supports  of  one  truss  are  not  directly  opposite  to  those  of 
the  other.  Tho  trusses  of  a  skew  bridge  are  usually  placed 
so  that  the  intermediate  panel  points  are  directly  opposite 
in  the  two  trusses,  and  the  floor  beams  arc  at  right  angles 
to  the  trusses.  When  the  skew  is  the  same  at  each  end  the 
trusses  are  symmefriml ;  otherwise  they  are  inis>immetriccd. 
In  the  analysis  of  unsymmetrical  trusses,  each  truss  must 
be  treated  separately  ;  and  the  stresses  are  to  be  computed 
for  all  the  members  of  the  truss. 

Frob.  142.   Fig.  C8  is  a  plan  and  Fig.  67  is  tlie  elevation 
of  one  of  the  two  trusses  of  an  unsymmetrical  through 


^ 


m 


it 


satwjiaiBBBwanBnw 


MISCELLANEOUS  TRUSSES. 


193 


9-10 

+  10.0 
+23.4 

+33.4 

I  the  end 
,o  tliose  of 
illy  placed 
y  opposite 
gilt  angles 
eh  end  the 
immetriccU. 
truss  must 
I  computed 

e  elevation 
il  through 


16 


Pratt  bridge;  the  span  2-18  is  120  feet^  *^«/«Pf  ^^^ 
feet  the  panels  3-5  and  2-4  are  each  18  feet,  the  panels 
ltl5  and  16-18  are  each  12  feet,  the  other  panels  are  ea.h 
15  feet,  the  inclination  of  the  -<!  P^^^^  ^^^l^'"  /"\t'i^^ 
the  same  as  that  of  the  diagonals  5-8  7-10,  ^^-^^^^  ^^^ 
etc  and  the  inclination  of  the  two  hip  verticals  3-4  and 
itio  s  the  same;  the  dead  and  live  loads  are  1000  lbs  a..d 
Im  lbs.  per  foot  per  truss:  find  all  the  stresses  in  all  the 

""m  first  find  the  chord  stresses  due  to  the  dead  load  as 

follows :         ' 

.  Dead  panel  load  at     4  =  U^  +  ^.5)  =  8.25  tons. 

''V        Dead  panel  load  at  16  =  ^  (6  +  7.5)  =  6.75  tons. 

:■:         Dead  panel  load  at  6,  8,  10,  12,  14  =  7  50  tons. 

Dead  load  reaction  at    2  =  25.5  tons. 
'         Dead  load  reaction  at  18  =  27.0  tons, 
tan  Z  3-2     with  vertical  =  .75  =  tan  Z 15-18  with  vertical, 
tan  Z  3-4      with  vertical  =  .15  =  tan  Z 15-16  with  vertical, 
tan  Z  3-6     with  vertica?  =  .9. 
tan  Z 14-15  with  vertical  =  .6. 
.      Heaceby(2)of  Art.  19  we  have  the  following  dead  load 

stresses: 

,  .    n  A  9K  K  V     75  =  19.1  tons. 

Dead  load  stress  m  2-4     =25.5X    .io 

•    AC      _  10  1  4-8  25  X  .15  =  20.3  tons. 
Dead  load  stress  in  4-6      =  19-1  +  »-^o  ^  •*" 

Dead  load  stress  in  16-18  =  27.0  x    .75  =  20.3  tons. 

Dead  load  stress  in  14-16  =  20.3  -  6.75  x  .16  =  19.3  tons, 
etc.  etc.  etc. 


i\ 


>  i. 


nmm 


il 


f  ( 


194 


HOOFS  AND   ItKIDGES. 


Also  the  following  greatest  live  load  stresses: 
Stress  in  3-6  =-■  ^i,  [13.6  x  12  +- 15  x  285]  1.345 
=  +  49.7  tons. 
Stress  in      5-G  =  -j^ll 02  +  15  x  198]  =  -  26.1  tons. 
Stress  in  12-13  =  ^^^  [297  +  15  x  222]  1.166  =  +  35.2  tons. 


Chord  Strebseh 

• 

MsMnEiw. 

2-4 

4-6 

+20.8 
+40.0 

6-S 

8-10 

10-12 

12-14 

U-16 

10-18 

7-11 

Demi  load  atreasos  .  .  . 
Ltvo  load  Btressea    .  .  . 

+l».l 
+:JSi.2 

+  85.  H 
+  71.fi 

+  43.1 

+  88.2 

+  41.0 
+  82.0 

+31.5 
+88.0 

+19.S 
+38.0 

+20.8 
+40.0 

-  44.8 

-  89.fi 

Maxliiiuin  strcsans  .  .  . 

+57.8 

+00.9 

+  107.4 

+  129.3 

+12.S.II 

+94..^ 

+57.9 

+00.9 

-184.4 

Stresses  in  tub  Diagonals. 


MKMBBR8. 

8-1 

8-0 

5-8 

7-10 

9-12 

11-14 

Dead  load  stresses . 
Live  load  st,resses  . 

+  8.3 
+  16.6 

+23.2 
+49.7 

+  12.2 
+35.1 

+  2.8 
+23.0 

-  6.6 

+  13.5 

-15.9 
+  6.3 

Maximum  stresses 

+24.0 

+  72.9 

+47.3 

+25.8 

+  6.9 

-  9.6 

Mkmukbs. 

15-10 

14-15 

12-18 

lO-ll 

8-9 

0-7  " 

Dead  load  stresses . 
Live  load  stresseo  . 

+  6.8 
+  13.6 

+  23.6 
+48.8 

+  15.9 
+35.2 

+  rt,6 
+23,9 

-  2.8 

+  14.7 

-12.2 

+  7.7 

Maximum  stresses . 

+20.4 

+  72.4 

+  51.1 

+30.6 

+  11.9 

-  4.r. 

We  see  from  the  above  that,  theoretically,  the  diagonals 
6-7  and  11-14  are  not  needed. 


I 


MKBter 


)45 

tons. 
;5.2  tons. 


lft-18 

7-11 

+20.8 
+40.6 

-  44.8 

-  89.6 

+60.9 

-184.4 

11-14 

6 

6 

-16.9 
+  6.3 

.9 

-  9.6 

6-7   ■ 

.8 

.7 

-12.2 

+  7.7 

.9 

-  4.f> 

.  i|i  ^^  j||i  ij^^iij^  I  .w^i jujj  < J»|Fi '^i  'I-  ■■  .14  i  ^  ^  I!"..  -V"  ■ ' .  ■■  '.■  ^  '^  *  ^.1 '  W! "  [  ■' 


iipwiLityi.ii.i'iy'^ppj 


MISCELLANEOUS  TliUSSES. 


Stresses  in  the  Posts. 


196 


MIMBKK8. 

2-3 

6-6 

7-S 

9-10 

11-12 

11-14 

16-18 

Dead  load  stresses  . 
Live  load  stresses  . 

-31.0 
-63.8 

-  0.8 
-26.1 

-  2.3 
-17.1 

-  0.0 
-12.6 

-  5.M 

-20.5 

-12.8 
-30.2 

-  33.8 

-  67.6 

Maximum  stresses . 

-05.7  -35.0 

-10.4 

-12.6 

-25.8 

-43.0 

-101.4 

Prob.  143.  Let  the  dimensions  of  Fig.  G7  be  as  follows: 
span  =  144  feet,  depth  =  24  feet,  the  panels  S-5  and  2-4 
=  21  feet,  the  panels  13-15  and  lG-18  =  15  feet,  all  the 
other  panels  =  18  feet ;  let  the  dead  and  live  loads  be 
800  lbs.  and  2000  lbs.  per  foot  per  truss:  find  all  the 
stresses  in  all  the  members. 


diagonals 


i 


